LeetCode 1531: String Compression II Solution

Master LeetCode problem 1531 (String Compression II), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1531. String Compression II

Hard

String Dynamic Programming

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming. We can define a 2D array dp[i][j] to represent the minimum length of the run-length encoded version of the substring ending at index i after deleting at most j characters. We can then iterate through the string and build this array by considering two cases: deleting the current character or keeping it.

At each step, we update the dp[i][j] based on these two cases. Finally, the answer will be the minimum value in the last row of the array.

The time complexity of this approach is O(n^2 * k) where n is the length of the input string and k is the maximum number of characters that can be deleted. The space complexity is O(n * k).

Solution Code

class Solution {
    public int getLengthOfOptimalCompression(String s, int k) {
        int n = s.length();
        int[][] dp = new int[n + 1][k + 1];
        
        for (int j = 1; j <= k; j++) {
            dp[0][j] = n; // Initialize to maximum possible value
        }
        
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j <= k; j++) {
                int delCount = 0, sameCount = 0;
                dp[i][j] = n;
                
                for (int prev = i; prev > 0 && delCount <= j; prev--) {
                    if (s.charAt(prev - 1) == s.charAt(i - 1)) {
                        sameCount++;
                        dp[i][j] = Math.min(dp[i][j], dp[prev - 1][j - delCount] + (sameCount < 2 ? 1 : String.valueOf(sameCount).length()));
                    } else {
                        delCount++;
                    }
                }
                
                dp[i][j] = Math.min(dp[i][j], dp[i - 1][j - 1]); // Delete current character
            }
        }
        
        return dp[n][k];
    }
}

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