You are given the two integers, n and m and two integer arrays, hBars and vBars. The grid has n + 2 horizontal and m + 2 vertical bars, creating 1 x 1 unit cells. The bars are indexed starting from 1. You can remove some of the bars in hBars from ho

maximizeSquareHoleArea.js

/**
 * @param {number} n
 * @param {number} m
 * @param {number[]} hBars
 * @param {number[]} vBars
 * @return {number}
 */
var maximizeSquareHoleArea = function(n, m, hBars, vBars) {
    /**
     * Finds the longest run of consecutive integers in an array.
     * Example: [2,3,4,7,8] → longest run = 3 (2→3→4)
     */
    function longestRun(arr) {
        // No removable bars → no run
        if (arr.length === 0) return 0;

        // Sort so we can detect consecutive sequences
        arr.sort((a, b) => a - b);

        let maxRun = 1;      // longest run found so far
        let currentRun = 1;  // current consecutive streak

        for (let i = 1; i < arr.length; i++) {
            // If the current bar continues the streak
            if (arr[i] === arr[i - 1] + 1) {
                currentRun++;
            } else {
                // Streak broken → reset
                currentRun = 1;
            }

            // Track the best streak seen
            maxRun = Math.max(maxRun, currentRun);
        }

        return maxRun;
    }

    // Longest consecutive removable horizontal bars
    const hRun = longestRun(hBars);

    // Longest consecutive removable vertical bars
    const vRun = longestRun(vBars);

    /**
     * IMPORTANT:
     * Removing k consecutive bars opens a gap of size (k + 1).
     * Example: remove bars 3 and 4 → you open 3 cells.
     */
    const hGap = hRun + 1;
    const vGap = vRun + 1;

    // The largest square must fit in BOTH directions
    const side = Math.min(hGap, vGap);

    // Area of the largest square hole
    return side * side;
};