LEETCODE RUNNING SUM OF 1D ARRAY (1 MAY- 7 MAY)

-

 class Solution {

    public int[] runningSum(int[] nums) {

        int index=0;

        int sum=0;

       for(int i=0;i<nums.length;i++){

           sum=sum+nums[i];

           nums[index]=sum;

           index++;

       }

        return nums;

        

    }

}

Comments

Post a Comment

Popular posts from this blog

Perfect Sum Problem Geeks for Geeks

-
 class Solution{ public int perfectSum(int arr[],int n, int sum)  {      int t[][]=new int[n+1][sum+1];     for(int i=0;i<n+1;i++){         for(int j=0;j<sum+1;j++){             if(i==0){                 t[i][j]=0;             }             if(j==0){                 t[i][j]=1;             }         }     }     int count=0;     for(int i=1;i<n+1;i++){         for(int j=1;j<sum+1;j++){             if(arr[i-1]>j){                 t[i][j]=t[i-1][j] % 1000000007;             }             el...
Read more

Array Formation HackerEarth

-
  import java . util .*; import java . util . LinkedList ; import java . util . Queue ;   class PrimeChecker { Queue < Integer > q ; Stack < Integer > stack ; Stack < Integer > stack1 ; public PrimeChecker (){ q = new LinkedList < Integer >(); stack = new Stack < Integer >(); } public void isPrime ( int num ){ boolean b = true ; for ( int i = 2 ; i <= num / 2 ; i ++) { int temp = num % i ; if ( temp == 0 ) { b = false ; break ; } } if ( b ){ //System.out.println(num + " is a Prime Number"); q . add ( num ); } else { //System.out.println(num + " is not a Prime Number"); stack . push ( num ); } } // This is checked so that we can skip // middle five numbers in below loop public void toPrint (){ //System.out.println(/*"Queue is: "+*/ q); //System.out.println(/*...
Read more

Recursive Sequence Geeks For Geeks Problem Of The Day 12-02-2024

-
Problem Link: https://www.geeksforgeeks.org/problems/recursive-sequence1611/1 Code: class Solution{     static long sequence(int n){                 long sum=0;         int pointer=0;                  int mod=1000000007;         int pointer2=1;                  for(int i=1;i<n+1;i++){            pointer=i;            long val=1;            int mul=pointer2;            for(int j=0;j<pointer;j++){                val=(val*mul)%mod;                mul++;            }            pointer2=mul;            sum=(sum+val)%mod; ...
Read more