Probability and Statistics [EXPLAINED]
IntuitionThe goal is to maximize the frequency of the most frequent element in an array by applying a set of operations. Each operation allows us to increase or
# Intuition
The goal is to maximize the frequency of the most frequent element in an array by applying a set of operations. Each operation allows us to increase or decrease any element by 1, and we can perform these operations up to k times. To achieve the maximum frequency, we need to find a way to optimize the application of these operations to clusters of similar elements in the array.

# Approach
Sort the array, then use a sliding window approach to track the current subarray that could form the most frequent element. For each possible size of the subarray, check if it's possible to make all elements in that subarray the same value with the given number of operations (k). The solution involves calculating the sum of differences within the window and adjusting it efficiently to ensure the number of operations stays within the limit.

# Complexity
- Time complexity:
O(n log n) due to the sorting of the array, where n is the length of the array. The sliding window itself runs in linear time O(n).

- Space complexity:
O(n) due to the storage used for the prefix sum array, which helps in efficiently calculating the sum of differences within the sliding window.

# Code
```typescript []
function maxFrequencyScore(a: number[], k: number): number {
    const n = a.length;

    a.sort((x, y) => x - y);

    const pf: number[] = Array(n + 1).fill(0);
    for (let i = 1; i <= n; i++) {
      pf[i] = pf[i - 1] + a[i - 1];
    }

    const pred = (m: number): boolean => {
      let possible = false;
      for (let i = m; i <= n; i++) {
        if (m % 2 === 1) {
          let deviation = 0;
          const ind = i - Math.floor(m / 2);
          const median = a[ind - 1];
          const s1 = pf[ind] - pf[i - m];

          deviation += (median * (Math.floor(m / 2) + 1)) - s1;

          const s2 = pf[i] - pf[ind];
          deviation += s2 - median * Math.floor(m / 2);

          if (deviation <= k) {
            possible = true;
          }
        } else {
          let ind1 = i - Math.floor(m / 2);
          let median1 = a[ind1 - 1];
          let deviation1 = 0;

          deviation1 += (median1 * Math.floor(m / 2)) - (pf[ind1] - pf[i - m]);
          deviation1 += (pf[i] - pf[ind1]) - median1 * Math.floor(m / 2);

          if (deviation1 <= k) {
            possible = true;
          }

          let ind2 = i - Math.floor(m / 2) + 1;
          let median2 = a[ind2 - 1];
          let deviation2 = 0;

          deviation2 += (pf[i] - pf[ind2 - 1]) - (median2 * Math.floor(m / 2));
          deviation2 += (median2 * Math.floor(m / 2)) - (pf[ind2 - 1] - pf[i - m]);

          if (deviation2 <= k) {
            possible = true;
          }
        }

        if (possible) {
          break;
        }
      }

      return possible;
    };

    let l = 0;
    let r = n + 1;
    while (r - l > 1) {
      const m = Math.floor((l + r) / 2);
      if (pred(m)) {
        l = m;
      } else {
        r = m;
      }
    }

    return l;
  }

```
['Array', 'Math', 'Binary Search', 'Sliding Window', 'Combinatorics', 'Binary Tree', 'Probability and Statistics', 'TypeScript']