Minimum operations to maximize last elements in arrays

Array Enumeration

43.9% acceptance

Feb 25, 2026

203

You are given two 0-indexed integer arrays, nums1 and nums2, both having length n. In an operation, you select an index i in the range [0, n - 1] and swap nums1[i] and nums2[i]. Find the minimum number of operations required to satisfy: nums1[n - 1] = max(nums1), nums2[n - 1] = max(nums2). Return the minimum number of operations needed, or -1 if it is impossible.

Solution

Rust

Time O(2^n)

Space O(n)

LeetCode

solution.rs

impl Solution {
  pub fn min_operations(nums1: Vec<i32>, nums2: Vec<i32>) -> i32 {
    let n = nums1.len();

    let solve = |m1: i32, m2: i32| -> Option<i32> {
      let mut ops = 0;
      for i in 0..n - 1 {
        let (a, b) = (nums1[i], nums2[i]);
        if a <= m1 && b <= m2 {
          // no swap needed
        } else if b <= m1 && a <= m2 {
          ops += 1; // swap
        } else {
          return None; // impossible
        }
      }
      Some(ops)
    };

    let m1 = nums1[n - 1];
    let m2 = nums2[n - 1];

    // Case 1: don't swap last
    let case1 = solve(m1, m2);

    // Case 2: swap last (cost 1 + solve with swapped maxes)
    let case2 = solve(m2, m1).map(|ops| ops + 1);

    match (case1, case2) {
      (Some(a), Some(b)) => a.min(b),
      (Some(a), None) => a,
      (None, Some(b)) => b,
      (None, None) => -1,
    }
  }
}