In a row of dominoes, tops[i] and bottoms[i] represent the top and bottom halves of the ith domino. (A domino is a tile with two numbers from 1 to 6 - one on each half of the tile.) We may rotate the ith domino, so that tops[i] and bottoms[i] swap v

minDominoRotations.js

/**
 * Finds the most frequently occurring number in an array.
 * @param {number[]} arr - The input array of numbers.
 * @return {Object} An object containing the most common item and its count.
 */
function getMostFrequentNumber(arr = []) {
    let mostFrequent = arr[0];
    let frequencyMap = {};

    for (let num of arr) {
        frequencyMap[num] = (frequencyMap[num] || 0) + 1;

        if (frequencyMap[mostFrequent] < frequencyMap[num]) {
            mostFrequent = num;
        }
    }

    return {
        item: mostFrequent,
        occurrences: frequencyMap[mostFrequent]
    };
}

var minDominoRotations = function (tops, bottoms) {
    // Find the most frequent values in each row
    const { item: mostFrequentTop, occurrences: countTop } = getMostFrequentNumber(tops);
    const { item: mostFrequentBottom, occurrences: countBottom } = getMostFrequentNumber(bottoms);

    // Choose the value that appears most frequently across both arrays
    const targetValue = countTop > countBottom ? mostFrequentTop : mostFrequentBottom;
    const primaryArray = countTop > countBottom ? tops : bottoms;
    const alternateArray = countTop <= countBottom ? tops : bottoms;

    let rotationCount = 0, n = primaryArray.length;

    // Check whether we can make all values equal to `targetValue`
    for (let i = 0; i < n; i++) {
        if (primaryArray[i] !== targetValue) {
            // If neither the primary nor alternate array contains the target value at this index, it's impossible
            if (alternateArray[i] !== targetValue) {
                return -1;
            } else {
                // Otherwise, we need to rotate this domino
                rotationCount++;
            }
        }
    }

    return rotationCount;
};