18. Subarray Sums Divisible by K

Topic :

arrays

Difficulty :

medium

Problem Link :

problem statement

Given an integer array nums and an integer k, return the number of non-empty subarrays that have a sum divisible by k.

A subarray is a contiguous part of an array.

Example 1:

Input: nums = [4,5,0,-2,-3,1], k = 5
Output: 7
Explanation: There are 7 subarrays with a sum divisible by k = 5:
[4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3]

Example 2:

Input: nums = [5], k = 9
Output: 0

Constraints:

1 <= nums.length <= 3 * 104
-104 <= nums[i] <= 104
2 <= k <= 104

solution

import java.io.*;
import java.util.*;
class Subarray_Sums_Divisible_by_K
{
    public static void main (String args[])
    {
        int nums[]={4,5,0,-2,-3,1};
        int k=5;
        System.out.println(subarraysDivByK(nums,k));
    }
     static int subarraysDivByK(int[] nums, int k) {
        
        HashMap <Integer,Integer> map=new HashMap<>();
        int ans=0,sum=0,rem=0;
        map.put(0,1);
        
        for(int i=0;i<nums.length;i++)
        {
            sum+=nums[i];
            
            rem=sum%k;
            
            if(rem<0)
                rem=rem+k;
            
            if(map.containsKey(rem))
            {
                ans+=map.get(rem);
                map.put(rem,map.get(rem)+1);
            }
            else
                map.put(rem,1);
        }
        
        return ans;
        
    }
}