Minimum cost to equalize arrays using swaps

Array Hash Table Greedy Counting

48.2% acceptance

Mar 31, 2026

You are given two integer arrays nums1 and nums2 of size n. You can perform the following two operations any number of times on these two arrays: Swap within the same array: Choose two indices i and j. Then, choose either to swap nums1[i] and nums1[j], or nums2[i] and nums2[j]. This operation is free of charge. Swap between two arrays: Choose an index i. Then, swap nums1[i] and nums2[i]. This operation incurs a cost of 1. Return an integer denoting the minimum cost to make nums1 and nums2 identical. If this is not possible, return -1.

Solution

Rust

Time O(n)

Space O(n)

LeetCode

solution.rs

use std::collections::HashMap;

impl Solution {
  pub fn min_cost(nums1: Vec<i32>, nums2: Vec<i32>) -> i32 {
    let n = nums1.len();
    let mut count = HashMap::new();
    let mut count1 = HashMap::new();
    
    for i in 0..n {
      if nums1[i] != nums2[i] {
        *count.entry(nums1[i]).or_insert(0i32) += 1;
        *count.entry(nums2[i]).or_insert(0i32) += 1;
        *count1.entry(nums1[i]).or_insert(0i32) += 1;
      }
    }
    
    if count.is_empty() {
      return 0;
    }
    
    for (_, &c) in &count {
      if c % 2 != 0 {
        return -1;
      }
    }
    
    let mut cost = 0;
    for (&v, &total) in &count {
      let have = *count1.get(&v).unwrap_or(&0);
      let need = total / 2;
      if have > need {
        cost += have - need;
      }
    }
    
    cost
  }
}