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.
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.
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.
Input: arr = [4,3,2,6,2,3,4], target = 6 Output: -1 Explanation: We have only one sub-array of sum = 6.
1 <= arr.length <= 105
1 <= arr[i] <= 1000
1 <= target <= 108
impl Solution {
pub fn min_sum_of_lengths(arr: Vec<i32>, target: i32) -> i32 {
let mut i = 0;
let mut sum = 0;
let mut pairs = vec![];
let mut ret = -1;
for j in 0..arr.len() {
sum += arr[j];
while i <= j && sum > target {
sum -= arr[i];
i += 1;
}
if sum == target {
match pairs.binary_search(&(i as i32, -1)) {
Err(0) | Ok(_) => (),
Err(k) => {
let x = pairs[k - 1].0 - pairs[k - 1].1 + (j - i) as i32 + 2;
if ret == -1 || ret > x {
ret = x;
}
}
}
let (a, b) = *pairs.last().unwrap_or(&(i32::MAX, 0));
if ((j - i) as i32) < a - b {
pairs.push((j as i32, i as i32));
}
}
}
ret
}
}