Minimum Operations to Make All Array Elements Equal - Sorting, Binary Search & Prefix Sum [JS]

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Description 

Solution: Sorting, Binary Search & Prefix Sum

  1. Sort nums in asc order.
  2. Get the prefix sum from the left and right of nums.
  3. For each query, binary search for the split point - the first element where nums[i] >= query
    From there, we can calculate the absolute difference of the two segments of nums.
    Segment one: query * i - leftSum[i - 1]
    Segment two: rightSum[i] - query * (n - i)

n = length of nums, m = number of queries
Time Complexity: O(n log(n) + m log(n))
Space Complexity: O(n + m)

var minOperations = function(nums, queries) {
  nums.sort((a, b) => a - b);
  let n = nums.length;
  let left = [...nums], right = [...nums];
  for (let i = 1; i < n; i++) left[i] += left[i - 1];
  for (let i = n - 2; i >= 0; i--) right[i] += right[i + 1];
  let ans = [];
  for (let query of queries) {
    let splitIndex = getSplitIndex(query);
    let leftDiff = splitIndex > 0 ? query * splitIndex - left[splitIndex - 1] : 0;
    let rightDiff = splitIndex < n ? right[splitIndex] - query * (n - splitIndex) : 0;
    ans.push(leftDiff + rightDiff);
  }
  return ans;
  
  function getSplitIndex(query) {
    let low = 0, high = n - 1;
    while (low < high) {
      let mid = Math.floor((low + high) / 2);
      if (nums[mid] >= query) high = mid;
      else low = mid + 1;
    }
    return nums[low] >= query ? low : n;
  }
};
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