#2025 - Maximum Number of Ways to Partition an Array

Problem Description

You are given a 0-indexed integer array nums of length n. The number of ways to partition nums is the number of pivot indices that satisfy both conditions:

1 <= pivot < n
nums[0] + nums[1] + ... + nums[pivot - 1] == nums[pivot] + nums[pivot + 1] + ... + nums[n - 1]

You are also given an integer k. You can choose to change the value of one element of nums to k, or to leave the array unchanged.

Return the maximum possible number of ways to partition nums to satisfy both conditions after changing at most one element.

Solution

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number}
 */
var waysToPartition = function(nums, k) {
  const n = nums.length;
  const prefixSums = [nums[0]];
  const leftDiffs = new Map();
  const rightDiffs = new Map();

  for (let i = 1; i < n; i++) {
    prefixSums[i] = prefixSums[i - 1] + nums[i];
    const diff = prefixSums[i - 1];
    rightDiffs.set(diff, (rightDiffs.get(diff) || 0) + 1);
  }

  const totalSum = prefixSums[n - 1];
  let result = totalSum % 2 === 0 ? (rightDiffs.get(totalSum / 2) || 0) : 0;

  for (let i = 0; i < n; i++) {
    const delta = k - nums[i];
    const newTotalSum = totalSum + delta;

    if (newTotalSum % 2 === 0) {
      const targetSum = newTotalSum / 2;
      const waysFromLeft = leftDiffs.get(targetSum) || 0;
      const waysFromRight = rightDiffs.get(targetSum - delta) || 0;
      result = Math.max(result, waysFromLeft + waysFromRight);
    }

    if (i < n - 1) {
      const currentDiff = prefixSums[i];
      leftDiffs.set(currentDiff, (leftDiffs.get(currentDiff) || 0) + 1);
      rightDiffs.set(currentDiff, rightDiffs.get(currentDiff) - 1);
    }
  }

  return result;
};