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Python
Bob code
def getLargestString(s, k):
frequency_array = [0] * 26
for i in range(len(s)):
frequency_array[ord(s[i]) -
ord('a')] += 1
ans = ""
i = 25
while i >= 0:
if (frequency_array[i] > k):
temp = k
st = chr( i + ord('a'))
while (temp > 0):
ans += st
temp -= 1
frequency_array[i] -= k
j = i - 1
while (frequency_array[j] <= 0 and
j >= 0):
j -= 1
if (frequency_array[j] > 0 and
j >= 0):
str1 = chr(j + ord( 'a'))
ans += str1
frequency_array[j] -= 1
else:
break
elif (frequency_array[i] > 0):
temp = frequency_array[i]
frequency_array[i] -= temp
st = chr(i + ord('a'))
while (temp > 0):
ans += st
temp -= 1
else:
i -= 1
return ans
if name == "main":
S = input()
k = 3
print (getLargestString(S, k))
Python
Bob code
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def count_special_strings(t, strings):
for s in strings:
count = 0
special_string_count = 0
for c in s:
if c == 'b' or c == '%':
count += 1
for i in range(1, len(s)):
if s[i:].count('b') + s[i:].count('%') == count:
special_string_count += 1
print(special_string_count)

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// N flowers on a Recatangular
int ans = 100000000;
void solve(vector<int> a, int n, int k, int index, int sum,
int maxsum)
{
if (k == 1)
{
maxsum = max(maxsum, sum);
sum = 0;
for (int i = index; i < n; i++)
{
sum += a[i];
}
maxsum = max(maxsum, sum);
ans = min(ans, maxsum);
return;
}
sum = 0;
for (int i = index; i < n; i++)
{
sum += a[i];
maxsum = max(maxsum, sum);
solve(a, n, k - 1, i + 1, sum, maxsum);
}
}
int GetMaxBeauty(int N, int K, vector<int> A)
{
solve(A, N, K, 0, 0, 0);
return ans;
}
// N flowers on a Recatangular 3PM Slots Ans

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//Given Array A having N integers and divisor K
int morethanNbyK(vector<int> arr, int n, int k)
{
int x = n / k;
int ans = 0;
unordered_map<int, int> freq;
for (auto i : arr)
freq[i]++;
for (auto i : freq)
{
if (i.second > x)
{
ans += i.first;
}
}
return ans;
}

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You are given Q queries. Each query contains a number N= Query[i] denoting the query
You have to find M such that :
* 1<=M<=N.
* M| M+1 | ... | N is a maximum as possible, where | is the OR bitwise operation
* M is as maximum possible .
* The answer to this query is the value of M.
Find the sum of answers to all queries modulo 10 pow 9 +7
Code please
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package com.ajit.String;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
int K = sc.nextInt();
String S = sc.next();
int result = solve(N, K, S);
System.out.println(result);
}
public static int solve(int N, int K, String S) {
int MOD = (int) Math.pow(10, 7);
Set<String> substrings = new HashSet<>();
for (int i = 0; i <= N - K; i++) {
substrings.add(S.substring(i, i + K));
}
int count = 0;
for (String substring : substrings) {
Set<String> permutations = new HashSet<>();
permute("", substring, permutations);
count += permutations.size();
count = count % MOD;
}
return count;
}
public static void permute(String prefix, String str, Set<String> permutations) {
int n = str.length();
if (n == 0) {
permutations.add(prefix);
} else {
for (int i = 0; i < n; i++) {
permute(prefix + str.charAt(i), str.substring(0, i) + str.substring(i + 1, n), permutations);
}
}
return;
}
}3PM Slots Ans

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Bob was Decorating

Wooden code

Rooted tree

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Max num of contests u can organise using bob
Code please
Nagapavan Yarramsetti:
You are given Q queries. Each query contains a number N= Query[i] denoting the query
You have to find M such that :
* 1<=M<=N.
* M| M+1 | ... | N is a maximum as possible, where | is the OR bitwise operation
* M is as maximum possible .
* The answer to this query is the value of M.
Code please
Find the sum of answers to all queries modulo 10 pow 9 +7
Solution:
import java.util.Scanner;
public class Main {
public static final int MOD = (int)1e9 + 7;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int T = sc.nextInt();
long sum = 0;
while (T-- > 0) {
int N = sc.nextInt();
int M = 0;
for (int i = 30; i >= 0; i--) {
if (((N >> i) & 1) == 1) {
M = (M << 1) + 1;
} else {
M = (M << 1);
}
}
sum = (sum + M) % MOD;
}
System.out.println(sum);
}
}
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Solution:
def count_bitwise_and_zero(nums: List[int]) -> int:
def get_set_bit_indices(x: int) -> List[int]:
"""Return indices of set bits in x"""
pow_2 = 1
exponent = 0
set_bits = []
while pow_2 <= x:
if pow_2 & x != 0:
set_bits.append(exponent)
exponent += 1
pow_2 *= 2
return set_bits
def is_bitwise_and_zero(window_length: int, bit_counts: Dict[int, int]) -> bool:
return all(value < window_length for value in bit_counts.values())
n = len(nums)
total_subarray_count = n * (n + 1) // 2
nonzero_subarray_count = 0
window_bit_counts = Counter()
"""At every iteration start, [left_idx, right_idx] is the longest subarray
ending at right_idx whose bitwise AND is nonzero."""
left_idx = 0
for right_idx, right_element in enumerate(nums):
if right_element == 0:
window_bit_counts.clear()
left_idx = right_idx + 1
continue
window_bit_counts += Counter(get_set_bit_indices(right_element))
while (left_idx < right_idx
and is_bitwise_and_zero(right_idx - left_idx + 1, window_bit_counts)):
window_bit_counts -= Counter(get_set_bit_indices(nums[left_idx]))
left_idx += 1
nonzero_subarray_count += (right_idx - left_idx + 1)
return total_subarray_count - nonzero_subarray_count

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Solution : array AR of size N

perfect_sum(arr, s, result) :
x = [0]*len(arr)
j = len(arr) – 1

while (s > 0) :
x[j] = s % 2
s = s // 2
j -= 1
sum = 0
for i in range(len(arr)) :
if (x[i] == 1) :
sum += arr[i]
if (sum == result) :
print(“{“,end=””);
for i in range(len(arr)) :
if (x[i] == 1) :
print(arr[i],end= “, “);
print(“}, “,end=””)

def print_subset(arr, K) :
x = pow(2, len(arr))
for i in range(1, x):
perfect_sum(arr, i, K)

# Driver code
arr = [ ]
n = int(input(“Enter length of array : “))
s=int(input(“Enter sum : “))
for i in range(n):
ele=int(input(“Enter element : “))
arr.append(ele)
print_subset(arr, s)

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Find all subarray index ranges in given Array with set bit sum equal to X

Given an array arr (1-based indexing) of length N and an integer X, the task is to find and print all index ranges having a set bit sum equal to X in the array.

Examples:

Input: A[] = {1 4 3 5 7}, X = 4

Output: (1, 3), (3, 4)

Explanation: In the above array subarray having set bit sum equal to X (= 4).

Starting from index 1 to 3. {1 4 3} = (001) + (100) + (011) = 4 and

other one is from 3 to 4 {3, 5} = (011) + (101) = 4.

Input: arr[] = {5, 3, 0, 4, 10}, X = 7

Output: (1 5)

Solution :

def countSetBit(arr, n):

    c = 0

    for i in range(n):

        x = arr[i]

        while (x):

            l = x % 10

            if (x & 1):

                c += 1

            x = x // 2

        arr[i] = c

        c = 0

def PrintIndex(arr, N, X, v):

    i,j,currSum = 0,0,arr[0]

    while (j < N and i < N):

        if (currSum == X):



            v.append(i + 1)

            v.append(j + 1)

            j += 1

            if(j<N):

                currSum += arr[j]

        elif (currSum < X):

            j += 1

            if(j<N):

                currSum += arr[j]

        else:

            currSum -= arr[i]

            i += 1

# Driver code

v = [1, 4, 3, 5, 7]

X = 4

N = len(v)

countSetBit(v, N)

ans = []

PrintIndex(v, N, X, ans)

for i in range(0,len(ans) - 1,2):

    print(f"({ans[i]} {ans[i + 1]})",end=" ")

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Soluion 1:

def findXor(arr,n):

    # Calculate xor of

    # all the elements

    xoR = 0;

    for i in range (0, n ) :

        xoR = xoR ^ arr[i]



    # Return twice of

    # xor value

    return xoR * 2

# Driver code

arr = [ 1, 5, 6 ]

n = len(arr)

print(findXor(arr, n))

Solution 2:

def findXorSum(arr, n):



    # variable to store the final Sum

    Sum = 0

    # multiplier

    mul = 1

    for i in range(30):

        c_odd = 0

        odd = 0

        for j in range(n):

            if ((arr[j] & (1 << i)) > 0):

                odd = (~odd)

            if (odd):

                c_odd += 1

        for j in range(n):

            Sum += (mul * c_odd)

            if ((arr[j] & (1 << i)) > 0):

                c_odd = (n - j - c_odd)

        mul *= 2

    return Sum

arr = [3, 8, 13]

n = len(arr)

print(findXorSum(arr, n))

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#include<iostream>
#include<unordered_map>
#include<string>
#include<math.h>
using namespace std;
int ans(string s){
unordered_map map;
for(int i =0; i map[s[i]]){
} }
return val; }
int main() {
string s;
cin>>s;
cout<< ans(s);
return 0; }
val = map[s[i]];
key = map[s[i]];

C++

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Solution :

def findMax(arr, n):

    large, index = 0, 0

    for i in range(n):

        if arr[i] > large:

            large = arr[i]

            index = i

    # return the index of the maximum element

    return index

def countSteps(arr, n):

     index = findMax(arr, n)

    steps = 1

    while index != 0:

        index = findMax(arr, index)

        steps += 1

    return steps

if __name__ == "__main__":

    arr = [2, 5, 8, 24, 4, 11, 6, 1, 15, 10]

    n = len(arr)

    print("Steps Taken:", countSteps(arr, n))

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Solution :

def findValue(arr, n):

    a = []

    b = []

    for i in range(n):

        a.append(arr[i] + i)

        b.append(arr[i] - i)

    x = a[0]

    y = a[0]

    for i in range(n):

        if (a[i] > x):

            x = a[i]

        if (a[i] < y):

            y = a[i]

    ans1 = (x - y)

    x = b[0]

    y = b[0]

    for i in range(n):

        if (b[i] > x):

            x = b[i]

        if (b[i] < y):

            y = b[i]

    ans2 = (x - y)

    return max(ans1, ans2)

if __name__ == '__main__':

    arr = [1, 2, 3, 1]

    n = len(arr)

    print(findValue(arr, n))

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Solution :

def getMaxToys(n,arr,money):

    L,H=0,n-1

    while(L<=H):

        if(sum(arr[L:H+1])<=money):

            break

        else:

            if arr[L]>arr[H]:

                L+=1

            else:

                H-=1

    return (H-L+1)



n=int(input())

arr=[]

for i in range(n):

    arr.append(int(input()))

money=int(input())

print(getMaxToys(n,arr,money))

max = 4

# Recursive function to find the

# required number of ways

def countWays(index, cnt, dp, n, m, k):

    # When all positions are filled

    if (index == n) :

        # If adjacent pairs are exactly K

        if (cnt == k):

            return 1

        else:

            return 0

# If already calculated

    if (dp[index][cnt] != -1):

        return dp[index][cnt]

   ans = 0

   # Next position filled with same color

    ans += countWays(index + 1, cnt, dp, n, m, k)

   # Next position filled with different color

    # So there can be m-1 different colors

    ans += (m - 1) * countWays(index + 1,

                               cnt + 1, dp, n, m, k)

   dp[index][cnt] = ans

    return dp[index][cnt]

# Driver Code

if __name__ == "__main__":

    n = 3

    m = 3

    k = 2

    dp = [[-1 for x in range(n + 1)]

              for y in range(max)]

   print(m * countWays(1, 0, dp, n, m, k))

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array AR of size N

perfect_sum(arr, s, result) :
x = [0]*len(arr)
j = len(arr) – 1

while (s > 0) :
x[j] = s % 2
s = s // 2
j -= 1
sum = 0
for i in range(len(arr)) :
if (x[i] == 1) :
sum += arr[i]
if (sum == result) :
print(“{“,end=””);
for i in range(len(arr)) :
if (x[i] == 1) :
print(arr[i],end= “, “);
print(“}, “,end=””)

def print_subset(arr, K) :
x = pow(2, len(arr))
for i in range(1, x):
perfect_sum(arr, i, K)

# Driver code
arr = [ ]
n = int(input(“Enter length of array : “))
s=int(input(“Enter sum : “))
for i in range(n):
ele=int(input(“Enter element : “))
arr.append(ele)
print_subset(arr, s)

Python

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public class XOR {
public static int findMaxOverlappingSegments(int[] A) {
int N = A.length;
int[][] dp = new int[N+1][N+1];
for (int i = 1; i <= N; i++) {
dp[i][i] = 0;
}
for (int i = 1; i <= N; i++) {
for (int j = i+1; j <= N; j++) {
dp[i][j] = -1;
}
}
for (int len = 2; len <= N; len++) {
for (int i = 1; i <= N-len+1; i++) {
int j = i+len-1;
int xor = 0;
for (int k = i; k <= j; k++) {
xor = xor ^ A[k-1];
}
if (xor == 0) {
dp[i][j] = Math.max(dp[i][j-1], dp[i][j-1] + dp[j][j] + 1);
} else {
dp[i][j] = dp[i][j-1];
}
}
}
return dp[1][N];
}
public static void main(String[] args) {
int[] A = {1, 2, 3, 4, 5, 6, 7, 8, 9};
System.out.println(findMaxOverlappingSegments(A));
}
}

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1.You have an array A of length N and two empty baskets
2.You are given an array Arr of size N
3.You have an weighted tree, rooted at vertex 1 that consists of N vertices and N-1 undirected edges it is given that we define path benefit as the minimum weight in the path

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Submask
Empty baskets
Non empty sub set
Smallest lexicographical array
Minimum poosibble cost
10^9+7
1)You are given an array A with permutation P with length N.
2) find total number of ways to participation A.
3) find the total number of elements in good term.
1. Find the lexicographical array possible after swapping
2.find the sum of F(i) for all where (1<=i<=N) and (I!=s). Since the answer might be large return it modulo 10^9+7

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import itertools
def permutation_cost(s):
# Get all possible permutations of the first 20 lowercase English letters
permutations = list(itertools.permutations(s))
# Set initial minimum cost to a large number
min_cost = float('inf')
for perm in permutations:
cost = 0
# Iterate through the permutation
for i in range(len(perm) - 1):
# If the next letter appears before the current letter in the permutation, add 1 to the cost
if perm[i] > perm[i + 1]:
cost += 1
# Update minimum cost if necessary
min_cost = min(min_cost, cost)
return min_cost
s = 'abcdefghijklmnopqrstuvwxyz'[:20]
print(permutation_cost(s))import itertools
def permutation_cost(s):
# Get all possible permutations of the first 20 lowercase English letters
permutations = list(itertools.permutations(s))
# Set initial minimum cost to a large number
min_cost = float('inf')
for perm in permutations:
cost = 0
# Iterate through the permutation
for i in range(len(perm) - 1):
# If the next letter appears before the current letter in the permutation, add 1 to the cost
if perm[i] > perm[i + 1]:
cost += 1
# Update minimum cost if necessary
min_cost = min(min_cost, cost)
return min_cost
s = 'abcdefghijklmnopqrstuvwxyz'[:20]
print(permutation_cost(s))

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2 baskets array solution!

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Smallest lexicographical array Minimum possible cost

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given an array A of size N.
You are allowed to choose at most one pair of elements such that distance (defined as the difference of their indices) is at most K and swap them.
Find the smallest lexicographical array possible after
Notes:
An array x is lexicographically smaller than an array y if there exists an index i such that xi <y i1 and x_{j} = y_{j} for all 0 <= j < i . Less formally, at the first index i in which they differ xi < yi
Input Formats@gman
The First-line contains Integers N Ea an integer, N, denoting the line i of the N subsequent lines (where describing A[i]. of elements in A. N) contains an integer
The next line contains an integer, K, denoting the upper bound on distance of index.
Constraints
Here as all the array values are equal swapping will not change the final result,
Here A=[5,4,3,2,11 K we can swap elements at index 0 and index 3 which makes A= [2,4,3,5,1].
Here A=[2,1,1,1,1] K we can swap elements at index 0 and index 3 chat which makes A= [1.1.1.2.11
bool swapped = false;
for (int i = 0; i < N - 1; i++) {
for (int j = i + 1; j <= min(i + K, N - 1); j++) {
if (A[i] > A[j]) {
swap(A[i], A[j]);
swapped = true;
break;
}
}
if (swapped) break;
}
if (!swapped) return A;
else return A;
C++

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Longest Subsquence

Python

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You are given a rooted tree of N vertices and an array A of N integers A[i] is the parent of the vertex (i+1 or 0 if the vertex (i+1) is the root.
For each vertex V from 1 to N, the answer K is the total number of subsets of vertices with the LCA (lowest common ancestor) equal to V Since the of K can be large calculate it modulo 10 ^ 9 + 7
Let there be an integer array Res where Res[i] contains the answer for (I + 1)th vertex, Find the array Res
Notes:
• The lowest common ancestor (LCA) is defined for X nodes A1, A2,.... Ax as the lowest node in the tree that has all A1 A2..... Ax as descendants (where we allow a node to be a descendant of itself)
Input Formate
The first line contains an integer N denoting the number of elements in A
Each line 1 of the N subsequent lines (where 0<=i<N) contains an integer
describing All It is given that A[i] denotes the parent of the vertex (i+1) or 0
If the vertex (i+1) is the root
const int N = 100005;
const int MOD = 1e9 + 7;
int a[N], dp[N], res[N];
vector<int> g[N];
void dfs(int u) {
dp[u] = 1;
for (int v : g[u]) {
dfs(v);
dp[u] = 1ll * dp[u] * (dp[v] + 1) % MOD;
}
res[u] = (1ll * dp[u] * (1 << g[u].size()) - 1 + MOD) % MOD;
}
vector<int> functionName(int n,vector<int>A)
{
g=A;
dfs(1);
return res;
}
Language c++
Infosys

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You are given an f length N and an array B of length M.
You need to find the longest possible sub-sequence (not necessarily of consecutive elements) of A such that it does not have amay B as a sub-arrey elements).
Find the length of the such sub-sequence.
A subsequence is a sequence that can be derived sequence by deleting some or no elements without changing the order of the • A subarray is a contiguous part of an array and maintains a relative ordering of elements.
Input Format
The first line contains an integer, N, denoting the size of array A.
Each line i of the N subsequent lines (where 0 ≤i≤N) contains an I integer describing A[i]
The next line contains an integer, M, denoting the size of array B.
Each line i of the M subsequent lines (where 0 ≤i≤ M) contains a an integer describing B[i].
Constraints
1 <= N <= 500
1 <= A[i] <= 100
1 <= M <= N
1 <= B[i] <= 100
N=4 A= [4, 4, 4. 4) M= 2 B= [4, 4] if you take any sub-sequence longer than 1, then array B will be the subarray of the chosen subsequence, so the answer here is 1.
N=5A [1, 2, 3,4,M =2 B=[6, 7] Since there is no common element between A and B, the longest subsequence answer is 5. we can choose is complete A hence the answer is 5
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N=5A= [1, 2, 3, 4, 5] M = 3 B = [1, 2, 3] we can not choose complete A since B will be a subarray of that so choose [1,2,4,5] as a subsequence and B is not a subarray of chosen subsequence hence the answer is 4.
func getMax (N int32, A []int32, M int32, B []int32) int32 { write your code here}

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def minJumps(arr, n):
jumps = [0 for i in range(n)]
if (n == 0) or (arr[0] == 0):
return float('inf')
jumps[0] = 0
for i in range(1, n):
jumps[i] = float('inf')
for j in range(i):
if (i <= j + arr[j]) and (jumps[j] != float('inf')):
jumps[i] = min(jumps[i], jumps[j] + 1)
break
return jumps[n-1]
arr = [1,1,1,2,4]
n = len(arr)
print('Minimum number of jumps to reach end is', minJumps(arr, n))

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#define ll long long int
void divideIt(int& no,int j,unordered_map<int,array<int,2>>& minFreq,
unordered_map<int,array<int,2>>& secondMinFreq, int idx){
int countFreq = 0 ;
while(no%j==0){
no /= j ;
countFreq++;
}
if( countFreq < minFreq[j][0] ){
secondMinFreq[j] = {minFreq[j][0],-1} ;
minFreq[j] = {countFreq,idx};
}
else if( countFreq == minFreq[j][0] )
secondMinFreq[j] = {minFreq[j][0],-1} ;
else if( countFreq > minFreq[j][0] && countFreq < secondMinFreq[j][0] )
secondMinFreq[j] = {countFreq,-1} ;
}
int calc(int N,int L,int R,vector<int> A,int Q,vector<int> B){
unordered_map<int, array<int,2> > minFreq;
unordered_map<int, array<int,2> > secondMinFreq;
unordered_map<int,int> countFreq ;
for(int i=L-1;i<R;i++){
int no = A[i] ;
if(no%2==0)
countFreq[2]++;
if(!minFreq.count(2))
minFreq[2] = {INT_MAX,int(-1)} ;
if(!secondMinFreq.count(2))
secondMinFreq[2] = {INT_MAX,int(-1)} ;
divideIt(no,2,minFreq,secondMinFreq,i) ;
for(int j=3;no>1;j+=2){
if(no%j==0)
countFreq[j]++;
if(!minFreq.count(j))
minFreq[j] = {INT_MAX,-1} ;
if(!secondMinFreq.count(j))
secondMinFreq[j] = {INT_MAX,-1} ;
divideIt(no,j,minFreq,secondMinFreq,i) ;
}
}
for(auto& e:countFreq){
if( e.second == R-L ){
if( minFreq[e.first][0] != 0 ){
for(int i=L-1;i<R;i++)
if( A[i]%e.first ){
minFreq[e.first] = {0,i} ;
}
}
}
else if( e.second < R-L ){
minFreq[e.first]={0,0} ;
secondMinFreq[e.first]={0,-1} ;
}
}
int countGood = 0 ;
for(auto e:B){
int minIdx = -1 ;
bool isP = true ;
int count2=0;
while(e%2==0){
e /=2 ;
count2++;
}
if( minFreq[2][0] < count2 )
minIdx = minFreq[2][1] ;
if(secondMinFreq[2][0]<count2)
isP=false;
for(int i=3;e>1;i+=2){
int countI = 0 ;
while(e%i==0){
e /= i ;
countI++;
}
if( secondMinFreq[i][0] < countI )
isP=false;
if( minFreq[i][0] < countI && minIdx!=-1 && minFreq[i][0] != minIdx )
isP = false;

__________

def longest_subsequence(A):
n = len(A)
dp = [1] * n # create an array to store the longest subsequence ending at each index
for i in range(1, n):
for j in range(i):
if A[j] > A[i] and dp[j] + 1 > dp[i]:
dp[i] = dp[j] + 1
return max(dp)

________

__________

perfect_sum(arr, s, result) :
x = [0]*len(arr)
j = len(arr) – 1
while (s > 0) :
x[j] = s % 2
s = s // 2
j -= 1
sum = 0
for i in range(len(arr)) :
if (x[i] == 1) :
sum += arr[i]
if (sum == result) :
print(“{“,end=””);
for i in range(len(arr)) :
if (x[i] == 1) :
print(arr[i],end= “, “);
print(“}, “,end=””)
def print_subset(arr, K) :
x = pow(2, len(arr))
for i in range(1, x):
perfect_sum(arr, i, K)
# Driver code
arr = [ ]
n = int(input(“Enter length of array : “))
s=int(input(“Enter sum : “))
for i in range(n):
ele=int(input(“Enter element : “))
arr.append(ele)
print_subset(arr, s)

___________

Infosys Latest Detailed Eligibility Criteria: Infosys SP and DSE

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How To Apply for Infosys Recruitment Drive 2022:

All candidates can apply for the Infosys Off-Campus Recruitment Drive 2022, they may apply for this position by clicking on the link given below before the link expire:

Join Our What's App Group

Step#1: Go to the above link, Join Our What's App Group.

Step#2: Apply Link: Click The Below Link To Apply

Apply Link: Click here (Recruitment Drive)
Preparation Materials Link: Click here (PDF Download)
Note: Apply for The Job Before Link Expires.
Infosys Off-Campus Drive 2022 – Frequently Asked Questions
How to Apply for Infosys Off-Campus 2022 Drive?
To apply for the Infosys Off Campus Drive, candidates must visit the official site at Infosys.com. Or else, check FreshersocJob.com to get the direct application link.
What is the Infosys Off-Campus Selection Process?
Online Written Test, Technical Interview, HR Interview.
Does FreshersocJob provide Infosys Job Updates?
Yes, FreshersocJob provides Infosys Job Updates.
Does FreshersocJob provide Infosys Placement Papers?
Yes, FreshersocJob provides Infosys Placement papers to find it under the FreshersocJob placement papers section
For More Detailed Information Regarding Placement Guides all Types of Updates Visit my Youtube Channel - Pappu Career Guide.

Also, Read More about this,

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Infor OFF Campus Drive For 2025 | 2024 | 2023 | 2022 | 2021 Batch | Associate Software Engineer Role | Any Graduates | Any Branch | Salary 6.5 LPA | Hyderabad, IN

Infosys 22nd January 2023 Both Slots 10 AM and 3 PM Live Coding Question and Solution | Refresh Every 5 Seconds for New Updates | SP | DSE Hiring | Page Vanish After the Exam

Infosys 22nd January 2023 Both Slots 10 AM and 3 PM Live Coding Question and Answers | Refresh Every 5 Seconds for New Updates | SP | DSE Hiring | Page Vanish After the Exam:

In this article, you can easily find the Infosys address, date of interview, eligibility details, interview syllabus, selection process, application link, and application procedure details explanation is given below.

In this article, you will get to know this drive, and also you can prefer a video that I created for your better understanding firstly understand the few points which I discussed in this video before applying.

For more such Latest new off-campus drives and for many years, internships, and free courses are updated on our website.

Infosys 2022 Recruitment Drive For Freshers – Any Batch:

Company Name: Infosys.
Experience: Freshers/Any

Job Title: Specialist Programmer and Digital Specialist Engineer
Country/Region: IN

Qualification: B.E/ B.Tech/ M.E/ M.Tech/ MCA/ M.Sc
Branch: Any Branch or Stream Eligible.

Job Location: PAN, India.

Year of Passing: 2022/ 2021/ 2020 Batch Only.

Stipend Details: INR ₹ 6,25,000 - 9,50,000 L.P.A.
Full-Time/Part-Time: Full-Time Employee
Application Deadline: Soon.
Job Posted: December 22, 2022.

Company Profile: Summary

Infosys Limited is an Indian multinational information technology company that provides business consulting, information technology and outsourcing services. The company was founded in Pune and is headquartered in Bangalore. Wikipedia

Get The Future You Want | www.Infosys.com

Infosys 22nd January 2023 Both Slots 3 PM Live Coding Question and Solution Refresh Every 5 Seconds for New Question or else use search Option 'Control F':

Compensation For Infosys Off-Campus Drive 2022:

The compensation for the Specialist Programmer role is INR 9.5 lakhs per annum and
for the Digital Specialist Engineer role it is INR 6.25 lakhs per annum

Infosys Latest Detailed Eligibility Criteria: Infosys SP and DSE:

Join Our Official Telegram Channel

Also, Read More about this,

Don't Miss Below Latest Job opportunity,

Also, Read the Latest Interview Experience,

TCS March Latest NQT Interview Experience 2022
Tips to Follow to Crack Tech Mahindra Exam 2022
CTS - Cognizant 96 Important Interview Questions 2022
TCS Ninja and Digital 2021 All Interview Questions 2022
Hexaware technical interview 2021

Stay tuned with www.freshersocjob.com to receive more such updates on recruitment. Join our Telegram Channel and Join our WhatsApp Group To get Instant updates for All Recruitment Internships and Off-Campus of 2021 Pass-Out Batch and Other Batch.

Comments

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Infosys 22nd January 2023 Both Slots 10 AM and 3 PM Live Coding Question and Solution | Refresh Every 5 Seconds for New Updates | SP | DSE Hiring | Page Vanish After the Exam

Infosys 22nd January 2023 Both Slots 10 AM and 3 PM Live Coding Question and Answers | Refresh Every 5 Seconds for New Updates | SP | DSE Hiring | Page Vanish After the Exam:

In this article, you can easily find the Infosys address, date of interview, eligibility details, interview syllabus, selection process, application link, and application procedure details explanation is given below.

In this article, you will get to know this drive, and also you can prefer a video that I created for your better understanding firstly understand the few points which I discussed in this video before applying.

For more such Latest new off-campus drives and for many years, internships, and free courses are updated on our website.

Infosys 2022 Recruitment Drive For Freshers – Any Batch:

Company Name: Infosys.Experience: Freshers/Any

Job Title: Specialist Programmer and Digital Specialist EngineerCountry/Region: IN

Qualification: B.E/ B.Tech/ M.E/ M.Tech/ MCA/ M.ScBranch: Any Branch or Stream Eligible.

Job Location: PAN, India. (adsbygoogle = window.adsbygoogle || []).push({});

Year of Passing: 2022/ 2021/ 2020 Batch Only.

Company Profile: Summary

Infosys Limited is an Indian multinational information technology company that provides business consulting, information technology and outsourcing services. The company was founded in Pune and is headquartered in Bangalore. Wikipedia

Get The Future You Want | www.Infosys.com

Infosys 22nd January 2023 Both Slots 3 PM Live Coding Question and Solution Refresh Every 5 Seconds for New Question or else use search Option 'Control F':

Compensation For Infosys Off-Campus Drive 2022:

Infosys Latest Detailed Eligibility Criteria: Infosys SP and DSE:

Also, Read More about this,

Stay tuned with www.freshersocjob.com to receive more such updates on recruitment. Join our Telegram Channel and Join our WhatsApp Group To get Instant updates for All Recruitment Internships and Off-Campus of 2021 Pass-Out Batch and Other Batch.

Comments

Post a Comment

Company Name: Infosys.
Experience: Freshers/Any

Job Title: Specialist Programmer and Digital Specialist Engineer
Country/Region: IN

Qualification: B.E/ B.Tech/ M.E/ M.Tech/ MCA/ M.Sc
Branch: Any Branch or Stream Eligible.

Job Location: PAN, India.