523. Continuous Subarray Sum

Problem Statement

Medium

Given an integer array nums and an integer k, return true if nums has a good subarray or false otherwise.

A good subarray is a subarray where:

its length is at least two, and
the sum of the elements of the subarray is a multiple of k.

Note that:

A subarray is a contiguous part of the array.
An integer x is a multiple of k if there exists an integer n such that x = n * k. 0 is always a multiple of k.

Example 1:

Input: nums = [23,2,4,6,7], k = 6
Output: true
Explanation: [2, 4] is a continuous subarray of size 2 whose elements sum up to 6.

Example 2:

Input: nums = [23,2,6,4,7], k = 6
Output: true
Explanation: [23, 2, 6, 4, 7] is an continuous subarray of size 5 whose elements sum up to 42.
42 is a multiple of 6 because 42 = 7 * 6 and 7 is an integer.

Example 3:

Input: nums = [23,2,6,4,7], k = 13
Output: false

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁹
0 <= sum(nums[i]) <= 2³¹ - 1
1 <= k <= 2³¹ - 1

C++

Source file

class Solution {
public:
    bool checkSubarraySum(vector<int>& nums, int k) {
        int running_sum = 0;
        for(int i=0; i<nums.size(); i++)
            nums[i] = running_sum += nums[i];
        unordered_map<int, int> seen;
        // Based on the result that if SUM_JmodK==SUM_ImodK  => (SUM_J-SUM_I)modK==0
        for(int i=0; i<nums.size(); i++){
            int rem = nums[i]%k;
            if(i && !rem)      // base case: contiguous subarray of all array elems upto index i
                return true;
            if(seen[rem]){
                if(i+1-seen[rem]>=2)  // result found
                    return true;
                // else dont update array, cuz it doesnt hurt to find the largest satisfying array
            }
            else seen[rem] = i+1;     // first occurance -> store
        }
        return false;   // not possible
    }
};