1838. Frequency of the Most Frequent Element - Detailed Explanation

Problem Statement

You are given an array of integers nums and an integer k. In one operation, you can choose an index and increment the element at that index by 1. The goal is to maximize the frequency of an element in the array after performing at most k operations.

The frequency of an element is defined as the number of times that element occurs in the array. Your task is to return the maximum possible frequency of an element after performing at most k operations.

Examples

Example 1

• Input:
  nums = [1, 2, 4], k = 5
• Output:
  3
• Explanation:
  We can perform the following operations:
  • Increase 1 by 3 → becomes 4 (using 3 operations)
  • Increase 2 by 2 → becomes 4 (using 2 operations)
    Now the array becomes [4, 4, 4] and the frequency of 4 is 3.

Example 2

• Input:
  nums = [1, 4, 8, 13], k = 5
• Output:
  2
• Explanation:
  No matter how you use the operations, you cannot equalize more than two numbers:
  • For instance, for window [1, 4]:
    Cost = (4*2) - (1+4) = 8 - 5 = 3 ≤ 5 → frequency 2.
  • For window [4, 8]:
    Cost = (8*2) - (4+8) = 16 - 12 = 4 ≤ 5 → frequency 2.
  • For window [8, 13]:
    Cost = (13*2) - (8+13) = 26 - 21 = 5 ≤ 5 → frequency 2. Expanding any window further would require more than 5 operations. Hence, the maximum frequency is 2.

Example 3

• Input:
  nums = [3, 9, 6], k = 2
• Output:
  1
• Explanation:
  After sorting, the array is [3, 6, 9].
  • For window [3, 6]:
    Cost = (6*2) - (3+6) = 12 - 9 = 3 > 2
  • For window [6, 9]:
    Cost = (9*2) - (6+9) = 18 - 15 = 3 > 2
    No window with length > 1 can be equalized within 2 operations, so the maximum frequency remains 1.

Constraints

• (1 \leq \text{nums.length} \leq 10^5)
• (1 \leq \text{nums}[i] \leq 10^5)
• (0 \leq k \leq 10^{10})

Hints

1. Sort the Array:
   Sorting the array allows you to work with elements in increasing order. This is key because when you try to “equalize” a group of numbers, you will always be raising the smaller numbers toward the larger ones.

2. Sliding Window Approach:
   Consider using two pointers (left and right) to maintain a window in the sorted array. For each window, compute the cost required to make every element in that window equal to the rightmost (largest) element.

   • Cost Formula:
     For a window from index l to r, the cost to make all elements equal to nums[r] is:
     [ \text{cost} = \text{nums}[r] \times (r - l + 1) - \text{sum(nums[l] ... nums[r])} ]
   • If the cost is within k, the window is valid, and you can try expanding it. Otherwise, move the left pointer to reduce the cost.

3. Update Maximum Frequency:
   Keep track of the maximum window size (number of elements that can be equalized) that satisfies the cost constraint.

Approach: Sliding Window

1. Sort the Array:
   Sorting the array ensures that the differences between consecutive elements are minimized, which helps in computing the cost efficiently.

2. Initialize Variables:

   • l = 0 (left pointer)
   • total = 0 to maintain the sum of elements in the current window
   • maxFreq = 1 to store the maximum frequency found

3. Expand the Window:
   For each index r (right pointer) from 0 to (n-1):

   • Add nums[r] to total.
   • Compute the current cost as:
     [ \text{cost} = \text{nums}[r] \times (r - l + 1) - \text{total} ]
   • If the cost is greater than k, move the left pointer l to the right while subtracting nums[l] from total until the cost is ≤ k.
   • Update maxFreq with the window size ((r - l + 1)) if it is greater than the current maxFreq.

4. Return maxFreq.

Step-by-Step Walkthrough

Consider nums = [1, 2, 4] and k = 5.

1. Sort the Array:
   Sorted nums = [1, 2, 4]

2. Initialize:

   • l = 0
   • total = 0
   • maxFreq = 1

3. Right Pointer r = 0:

   • Add nums[0] → total = 1
   • Window size = 1, cost = (1 \times 1 - 1 = 0) ≤ k
   • Update maxFreq = 1

4. Right Pointer r = 1:

   • Add nums[1] → total = 1 + 2 = 3
   • Window size = 2, cost = (2 \times 2 - 3 = 4 - 3 = 1) ≤ k
   • Update maxFreq = 2

5. Right Pointer r = 2:

   • Add nums[2] → total = 3 + 4 = 7
   • Window size = 3, cost = (4 \times 3 - 7 = 12 - 7 = 5) ≤ k
   • Update maxFreq = 3

Result: Maximum frequency = 3.

Code Implementation

Python Code