Maximum Frequency After Subarray Operation: An In-depth Solution | LeetCode

In this blog post, we will explore a common algorithmic problem, Maximum Frequency After Subarray Operation. It’s an interesting problem that challenges your understanding of arrays and frequency calculations, which are essential concepts for anyone learning data structures and algorithms.

Problem Overview

You are given an array nums of length n and an integer k. The goal is to find the maximum frequency of the value k after performing an operation on a subarray of nums. The operation involves selecting a contiguous subarray and adding a value x to each element of that subarray. After performing this operation once, the task is to determine how many times k appears in the modified array.

To better understand the problem, let’s go over a couple of examples.

Example 1:

Input:
nums = [1, 2, 3, 4, 5, 6], k = 1

Output:
2

Explanation:
By adding -5 to the subarray nums[2..5], we get the array [1, 2, -2, -1, 0, 1]. After the operation, the number 1 appears twice.

Example 2:

Input:
nums = [10, 2, 3, 4, 5, 5, 4, 3, 2, 2], k = 10

Output:
4

Explanation:
By adding 8 to the subarray nums[1..9], we get the array [10, 10, 11, 12, 13, 13, 12, 11, 10, 10]. After the operation, the number 10 appears four times.

To solve this problem efficiently, we need to focus on how to maximize the frequency of the value k after the operation. Let’s break down the steps involved in achieving this.

Step 1: Understand the Subarray Operation

The core of this problem is performing the operation on a subarray. A subarray is simply a contiguous section of the array. By selecting a subarray, we can modify all of its elements by adding a fixed number x. The goal is to choose the right subarray and the correct value of x such that the frequency of k in the modified array is maximized.

Step 2: Using a Sliding Window Approach

A sliding window technique works well for this type of problem because it allows us to efficiently explore all possible subarrays without the need to check each one individually. The idea is to maintain a window over the array where we add x to the elements in that window and calculate the resulting frequency of k.

In our approach, the key is to maximize the number of ks after applying the operation. This involves iterating through all possible values for x and determining which results in the highest frequency for k.

Step 3: Algorithm Walkthrough

We can break the algorithm down into the following steps:

1. Count the occurrences of k: First, we calculate how many times k appears in the original array. This gives us a baseline to start with.

2. Simulate the effect of adding x to each subarray: For each potential subarray and corresponding value of x, calculate how the frequency of k changes. We’ll use a sliding window technique to efficiently explore all possible subarrays.

3. Track the maximum frequency: Keep track of the highest frequency of k encountered during the operation.

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The Code Solution

#include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
    int maxFrequency(vector<int>& nums, int k) {
        int total_k = 0;
        // Count the occurrences of k in the original array
        for (auto num : nums) {
            if (num == k) {
                total_k++;
            }
        }
        vector<int> nerbalithy = nums;  // Create a copy of nums
        int max_gain = 0;
        // Try all possible values for x to add to subarrays
        for (int v = 1; v <= 50; v++) {
            if (v == k) continue;
            
            int x = k - v;
            int current_sum = 0;
            int local_max = 0;
            // Sliding window approach to simulate the subarray operation
            for (auto num : nums) {
                if (num == v) {
                    current_sum += 1;
                } else if (num == k) {
                    current_sum -= 1;
                }
                
                if (current_sum < 0) {
                    current_sum = 0;  // Reset when sum is negative
                }
                
                if (current_sum > local_max) {
                    local_max = current_sum;  // Update the local max
                }
            }
            
            if (local_max > max_gain) {
                max_gain = local_max;  // Track the maximum gain
            }
        }
        
        int max_freq = total_k + max_gain;  // Calculate the final frequency of k
        return max_freq;
    }
};
Explanation of the Code:

1. Initial Count of k: The first loop counts the number of times k appears in the array nums.

2. Sliding Window Technique: We loop through each possible value for v, excluding k, and calculate the frequency changes when we add x = k - v to the subarray.

3. Tracking Maximum Gain: For each value of v, we track the local maximum gain in frequency after applying the operation to the subarray. The final result is the sum of the original occurrences of k and the best possible gain from the subarray operation.

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Time Complexity

The time complexity of the solution is O(n * 50) because we are iterating through the array once for each value of v (where v ranges from 1 to 50). Since n can go up to 10^5, this approach efficiently handles the upper limits of the problem's constraints.