#2499 - Minimum Total Cost to Make Arrays Unequal

Problem Description

You are given two 0-indexed integer arrays nums1 and nums2, of equal length n.

In one operation, you can swap the values of any two indices of nums1. The cost of this operation is the sum of the indices.

Find the minimum total cost of performing the given operation any number of times such that nums1[i] != nums2[i] for all 0 <= i <= n - 1 after performing all the operations.

Return the minimum total cost such that nums1 and nums2 satisfy the above condition.

In case it is not possible, return -1.

Solution

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number}
 */
var minimumTotalCost = function(nums1, nums2) {
  const n = nums1.length;
  const conflictIndices = [];
  const valueCount = new Map();

  for (let i = 0; i < n; i++) {
    if (nums1[i] === nums2[i]) {
      conflictIndices.push(i);
      valueCount.set(nums1[i], (valueCount.get(nums1[i]) || 0) + 1);
    }
  }

  if (conflictIndices.length === 0) return 0;

  let dominantValue = -1;
  let maxCount = 0;

  for (const [value, count] of valueCount) {
    if (count > maxCount) {
      maxCount = count;
      dominantValue = value;
    }
  }

  const swapsNeeded = conflictIndices.length;
  let helpersNeeded = Math.max(0, 2 * maxCount - swapsNeeded);

  const helperIndices = [];
  for (let i = 0; i < n && helpersNeeded > 0; i++) {
    if (nums1[i] !== nums2[i] && nums1[i] !== dominantValue && nums2[i] !== dominantValue) {
      helperIndices.push(i);
      helpersNeeded--;
    }
  }

  if (helpersNeeded > 0) return -1;

  const allIndices = [...conflictIndices, ...helperIndices];
  allIndices.sort((a, b) => a - b);

  let result = 0;
  for (let i = 0; i < allIndices.length; i++) {
    result += allIndices[i];
  }

  return result;
};