#include<bits/stdc++.h>
using namespace std;

long long m = 100000000000000000;

int minOperations(vector<int> &arr)
{
    int n=arr.size();
    if(n == 1) return 0;

    if(n == 2) return arr[0]+arr[1]<0;

    int cnt = 0;

    long long prev,curr,next;

    prev = m;
    curr = arr[0];
    next = arr[1];

    int ind = 1;

    while(ind<n)
    {   
        if(prev+curr+next<0 || curr+next<0)
        {
            prev = curr;
            curr = m;
            cnt++;
        }else{
            prev = curr;
            curr = next;
        }

        if(ind<n) next = arr[ind+1];
        ind++;
    }

    return cnt;
}

UKG ✅
Make the Array Positive

def find_consecutive_sets(n):
    result = []
    count = 0

    for i in range(1, n//2 + 2):
        sum_ = 0
        temp = []
        
        for j in range(i, n + 1):
            sum_ += j
            temp.append(j)
            
            if sum_ == n:
                result.append(temp[:]) 
                count += 1
                break
            elif sum_ > n:
                break

    result.reverse()

    for seq in result:
        print(" ".join(map(str, seq)))
    
    print(count)

n = int(input())
find_consecutive_sets(n)

Deloitte ✅

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#include <bits/stdc++.h>
using namespace std;
long long getMaximumSum(vector<int>& no_adjacent, vector<int>& one_adjacent, vector<int>& both_adjacent) {
    int n = no_adjacent.size();
    vector<vector<long long>> dp(n, vector<long long>(3, 0));
    dp[0][0] = no_adjacent[0];
    dp[0][1] = one_adjacent[0];
    dp[0][2] = 0; 
        for (int i = 1; i < n; ++i) {
        dp[i][0] = max({dp[i - 1][0], dp[i - 1][1], dp[i - 1][2]}) + no_adjacent[i];
        dp[i][1] = max(dp[i - 1][0], dp[i - 1][2]) + one_adjacent[i];
        dp[i][2] = dp[i - 1][1] + both_adjacent[i];
    }
    
    return max({dp[n - 1][0], dp[n - 1][1], dp[n - 1][2]});
}

Meesho ✅

def ways(total, k):
    MOD = 1000000007
    dp = [0] * (total + 1)
    dp[0] = 1
    for i in range(1, k + 1):
        for w in range(i, total + 1):
            dp[w] = (dp[w] + dp[w - i]) % MOD
    return dp[total]

Meesho ✅

int beauty( const vector<int> &A )
{
   vector<int> B( 1 + A.size(), 0 );
   int N = A.size();
   for ( int pos = 1; pos <= N; pos++ )
   {
      auto B_old = B;
      int a = A[pos-1];      // index is (position - 1)
      for ( int length = 1; length <= pos; length++ )
      {
         B[length] = max( B_old[length-1] + ( a == length ), B_old[length] );
      }
   }

   return *max_element( B.begin(), B.end() );
}

Meesho ✅

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- Fresh B. Tech Graduates with Knowledge of Cyber Security
- Good communication skills
- Willing to Work from Office / Open to Shifts

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Nike is hiring Software Engineer

For 2021, 2022, 2023 gards
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Interested candidate send your resume this mail id

nandini.sharma@exlservice.com

TGH Software Hiring Software Developer

Batch : 2020/21/22/23

https://docs.google.com/forms/d/e/1FAIpQLSfrRM3v4YFkLqL4Z_SANg6sdI38AC-tz0D49sTYvD706Q1ubQ/viewform

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int[][] tiles = new int[n][n];
        
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                tiles[i][j] = scanner.nextInt();
            }
        }
        
        System.out.println(solve(n, tiles));
        scanner.close();
    }
    
    public static int solve(int n, int[][] tiles) {
        int[][] dp2 = new int[n][n];
        int[][] dp5 = new int[n][n];
        
        int[][] factors2 = new int[n][n];
        int[][] factors5 = new int[n][n];
        
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                factors2[i][j] = countFactor(tiles[i][j], 2);
                factors5[i][j] = countFactor(tiles[i][j], 5);
            }
        }
        
        dp2[0][0] = factors2[0][0];
        dp5[0][0] = factors5[0][0];
        
        for(int j = 1; j < n; j++) {
            dp2[0][j] = dp2[0][j-1] + factors2[0][j];
            dp5[0][j] = dp5[0][j-1] + factors5[0][j];
        }
        
        for(int i = 1; i < n; i++) {
            dp2[i][0] = dp2[i-1][0] + factors2[i][0];
            dp5[i][0] = dp5[i-1][0] + factors5[i][0];
        }
        
        for(int i = 1; i < n; i++) {
            for(int j = 1; j < n; j++) {
                dp2[i][j] = Math.min(dp2[i-1][j], dp2[i][j-1]) + factors2[i][j];
                dp5[i][j] = Math.min(dp5[i-1][j], dp5[i][j-1]) + factors5[i][j];
            }
        }
        
        return Math.min(dp2[n-1][n-1], dp5[n-1][n-1]);
    }
    
    public static int countFactor(int num, int factor) {
        if(num == 0) return Integer.MAX_VALUE;
        int count = 0;
        while(num % factor == 0) {
            count++;
            num /= factor;
        }
        return count;
    }
}

Treasure Room ✅
Java

#include <bits/stdc++.h>
using namespace std;
int solve(string s)
{
    if (s.empty()) 
        return 0;
    int n = s.size();
    char lastchar = s[s.size() - 1];
    char firstchar = s[0];
    vector<int> indexlast;
    for (int i = 0; i < n; i++) {
        if (s[i] == lastchar)
            indexlast.push_back(i);
    }

    int mini = INT_MAX;
    for (int i = 0; i < n; i++) {
        if (s[i] == firstchar) {
            int index = *lower_bound(indexlast.begin(), indexlast.end(), i);
            mini = min(mini, index - i + 1);
        }
    }

    return n - mini;
}

Amazon ✅