18. Find Two Non-overlapping Sub-arrays Each With Target Sum

Given an array of integers arr and an integer target.

You have to find two non-overlapping sub-arrays of arr each with a sum equal target. There can be multiple answers so you have to find an answer where the sum of the lengths of the two sub-arrays is minimum.

Return the minimum sum of the lengths of the two required sub-arrays, or return -1 if you cannot find such two sub-arrays.

Example 1:

Input: arr = [3,2,2,4,3], target = 3
Output: 2
Explanation: Only two sub-arrays have sum = 3 ([3] and [3]). The sum of their lengths is 2.

Example 2:

Input: arr = [7,3,4,7], target = 7
Output: 2
Explanation: Although we have three non-overlapping sub-arrays of sum = 7 ([7], [3,4] and [7]), but we will choose the first and third sub-arrays as the sum of their lengths is 2.

Example 3:

Input: arr = [4,3,2,6,2,3,4], target = 6
Output: -1
Explanation: We have only one sub-array of sum = 6.

Example 4:

Input: arr = [5,5,4,4,5], target = 3
Output: -1
Explanation: We cannot find a sub-array of sum = 3.

Example 5:

Input: arr = [3,1,1,1,5,1,2,1], target = 3
Output: 3
Explanation: Note that sub-arrays [1,2] and [2,1] cannot be an answer because they overlap.

Solution: (Prefix and Suffix sum)

Approach: Finding Prefix sum and getting min length of all possible subarrays at every index Finding Suffix sum and getting min length of all possible subarrays at every index Finding min length of two subarray at index : len = ps[0 - i-1] + ss[i - n]

class Solution
{
public:
    int minSumOfLengths(vector<int> &arr, int target)
    {

        int n = arr.size();

        vector<int> ps(n);
        unordered_map<int, int> pidx;
        pidx[0] = -1;
        int s = 0;
        int minArr = INT_MAX;

        for (int i = 0; i < arr.size(); i++)
        {
            s += arr[i];

            if (pidx.find(s - target) != pidx.end())
            {
                int dist = i - pidx[s - target];
                minArr = min(minArr, dist);
            }

            ps[i] = minArr;
            pidx[s] = i;
        }

        vector<int> ss(n);
        unordered_map<int, int> sidx;
        minArr = INT_MAX;
        s = 0;
        sidx[0] = arr.size();

        for (int i = arr.size() - 1; i >= 0; i--)
        {
            s += arr[i];

            if (sidx.find(s - target) != sidx.end())
            {
                int dist = sidx[s - target] - i;
                minArr = min(minArr, dist);
            }

            ss[i] = minArr;
            sidx[s] = i;
        }

        int res = INT_MAX;

        for (int i = 1; i < n; i++)
        {

            if (ps[i - 1] != INT_MAX && ss[i] != INT_MAX)
            {
                res = min(res, ps[i - 1] + ss[i]);
            }
        }

        if (res == INT_MAX)
        {
            return -1;
        }

        return res;
    }
};

Time Complexity: O(n)