Array     C++     Hash Table     Medium     Prefix Sum    

Problem Statement:

Given an array of positive integers nums, remove the smallest subarray (possibly empty) such that the sum of the remaining elements is divisible by p. It is not allowed to remove the whole array.

Return the length of the smallest subarray that you need to remove, or -1 if it's impossible.

A subarray is defined as a contiguous block of elements in the array.

 

Example 1:

Input: nums = [3,1,4,2], p = 6
Output: 1
Explanation: The sum of the elements in nums is 10, which is not divisible by 6. We can remove the subarray [4], and the sum of the remaining elements is 6, which is divisible by 6.

Example 2:

Input: nums = [6,3,5,2], p = 9
Output: 2
Explanation: We cannot remove a single element to get a sum divisible by 9. The best way is to remove the subarray [5,2], leaving us with [6,3] with sum 9.

Example 3:

Input: nums = [1,2,3], p = 3
Output: 0
Explanation: Here the sum is 6. which is already divisible by 3. Thus we do not need to remove anything.

 

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= nums[i] <= 109
  • 1 <= p <= 109

Solution:

Basically the solution is to find a subarray whose mod equals that of the complete array. Let us call it M. The first task is to find M.

 
int M = 0;
for(int num: nums) {M += num; M %= p;}

Now we need to find subarray whose mod equals M. To do this we use a logic similar to Two Sum. The logic is that while traversing the array, at any position i we try to find a complement at an earlier index j in the [0,i) range.

Say the prefix sum modulus at j is a variable a and the prefix sum modulus at i is a+M. Then the modulus of the subarray is M and our task is done. Now as we traverse we are going to search for the variable a assuming the current modulus is a+M.

So, say at index i, the current prefix sum modulus is mod. Then we need to find an index j where the prefix sum modulus is mod-M. But this can be negative. So we can just use (mod-M+p) % p.

To find this complement, we use the exact same logic of Two Sum ie using a hash map.

 cpp []
class Solution {
public:
    int minSubarray(vector<int>& nums, int p) 
    {
        int M = 0;
        for(int num: nums) {M += num; M %= p;}
        if(M==0) return 0;

        // Now we need to find smallest subarray whose mod is equal to M
        int n = nums.size(), mod = 0, res = INT_MAX;
        unordered_map<int,int> mods;
        for(int i=0; i<n; i++)
        {
            int num = nums[i];
            mod += num; mod %= p;
            int complement = (mod+p-M) % p;

            if (mods.count(complement)) res = min(res, i-mods[complement]);
            if (complement==0) res = min(res, i+1);
            mods[mod] = i;
        }
        if (res==INT_MAX || res==n) return -1;
        return res;
    }
};