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1477. Find Two Non-overlapping Sub-arrays Each With Target Sum

You are 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. 

Constraints:

• 1 <= arr.length <= 105
• 1 <= arr[i] <= 1000
• 1 <= target <= 108

Solutions (Rust)

1. Solution

implSolution{pubfnmin_sum_of_lengths(arr:Vec<i32>,target:i32) -> i32{letmut i = 0;letmut sum = 0;letmut pairs = vec![];letmut ret = -1;for j in0..arr.len(){ sum += arr[j];while i <= j && sum > target { sum -= arr[i]; i += 1;}if sum == target {match pairs.binary_search(&(i asi32, -1)){Err(0) | Ok(_) => (),Err(k) => {let x = pairs[k - 1].0 - pairs[k - 1].1 + (j - i)asi32 + 2;if ret == -1 || ret > x { ret = x;}}}let(a, b) = *pairs.last().unwrap_or(&(i32::MAX,0));if((j - i)asi32) < a - b { pairs.push((j asi32, i asi32));}}} ret }}