1477. Find Two Non-overlapping Sub-arrays Each With Target Sum
Medium
Given an array of integers arr
and an integer target
.
You have to find two non-overlapping sub-arrays of arr
each with 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.
Constraints:
1 <= arr.length <= 10^5
1 <= arr[i] <= 1000
1 <= target <= 10^8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 | class Solution { public int minSumOfLengths(int[] arr, int target) { Map<Integer, Integer> map = new HashMap<>(); map.put(0,-1); int n = arr.length; int sum = 0; for(int i=0; i<n; ++i){ sum += arr[i]; map.put(sum, i); } sum = 0; int minLeft = Integer.MAX_VALUE; int min = Integer.MAX_VALUE; for(int i=0; i<n; ++i){ sum += arr[i]; if(map.containsKey(sum-target)){ //left subarr exists minLeft = Math.min(minLeft, i-map.get(sum-target)); } if(minLeft != Integer.MAX_VALUE && map.containsKey(sum+target)){ min = Math.min(min, minLeft+map.get(sum+target)-i); } } return min==Integer.MAX_VALUE?-1:min; } } |
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