LeetCode 3505: Minimum Operations to Make Elements Within K Subarrays Equal Solution

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3505. Minimum Operations to Make Elements Within K Subarrays Equal

Hard

Array Hash Table Math Dynamic Programming Sliding Window Heap (Priority Queue)

Problem Explanation

Explanation:

To solve this problem, we can use a sliding window approach. We need to find the minimum number of operations required to make at least k non-overlapping subarrays of size x in the given array, such that all elements within each subarray are equal.

First, we calculate the frequency of each unique element in the array.
Then, we iterate through the array with a sliding window of size x, keeping track of the total operations needed to make all elements in that window equal to the most frequent element in that window.
We update the total operations as we slide the window along the array.
Finally, we return the total operations needed to achieve at least k non-overlapping subarrays.

Solution Code

class Solution {
    public int minOperations(int[] nums, int x, int k) {
        Map<Integer, Integer> freqMap = new HashMap<>();
        for (int num : nums) {
            freqMap.put(num, freqMap.getOrDefault(num, 0) + 1);
        }

        int totalOps = 0;
        int opsInWindow = 0;
        int maxFreq = 0;

        for (int i = 0; i < x * k; i++) {
            int freq = freqMap.get(nums[i]);
            maxFreq = Math.max(maxFreq, freq);
            opsInWindow += Math.abs(maxFreq - freq);
        }

        totalOps = opsInWindow;

        for (int i = x * k; i < nums.length; i++) {
            int outNum = nums[i - x * k];
            int inNum = nums[i];

            freqMap.put(outNum, freqMap.get(outNum) - 1);
            freqMap.put(inNum, freqMap.getOrDefault(inNum, 0) + 1);

            maxFreq = Math.max(maxFreq, freqMap.get(inNum));
            opsInWindow += Math.abs(maxFreq - freqMap.get(inNum));

            freqMap.put(outNum, freqMap.get(outNum) + 1);
            if (freqMap.get(outNum) == 0) {
                freqMap.remove(outNum);
                maxFreq = 0;
                for (int num : freqMap.values()) {
                    maxFreq = Math.max(maxFreq, num);
                }
            }

            opsInWindow -= Math.abs(maxFreq - freqMap.get(outNum));

            totalOps = Math.min(totalOps, opsInWindow);
        }

        return totalOps;
    }
}

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