Minimum Deletions

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Time: O(n*m)

Space: O(n*m)

Intuition

Minimum deletions to make string have distinct character frequencies. Sort frequencies, greedily decrement.

Algorithm

1Count frequencies. Sort descending. For each freq: if already used: decrement until unique or 0.

Common Pitfalls

•Same as LC 1647. Use a set of used frequencies. Greedy: reduce until you find unused frequency or 0.

Minimum Deletions.java

Java

// Approach: LCS-based. Min deletions = (len(s1) - LCS) + (len(s2) - LCS).
// Time: O(n*m) Space: O(n*m)
class Solution {
    static int minDeletions(String s) {
        int n = s.length();
        String rev = new StringBuilder(s).reverse().toString();
        return n - longestCommonSubsequence(s, rev);
    }

    static int longestCommonSubsequence(String s1, String s2) {
        int n = s1.length();
        int[][] dp = new int[n + 1][n + 1];

        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                if (s1.charAt(i - 1) == s2.charAt(j - 1))
                    dp[i][j] = 1 + dp[i - 1][j - 1];
                else
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
            }
        }

        return dp[n][n];
    }
}

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