LeetCode //C - 1209. Remove All Adjacent Duplicates in String II

1209. Remove All Adjacent Duplicates in String II

You are given a string s and an integer k, a k duplicate removal consists of choosing k adjacent and equal letters from s and removing them, causing the left and the right side of the deleted substring to concatenate together.

We repeatedly make k duplicate removals on s until we no longer can.

Return the final string after all such duplicate removals have been made. It is guaranteed that the answer is unique.

Example 1:

Input: s = "abcd", k = 2

Output: "abcd"

Explanation: There's nothing to delete.

Example 2:

Input: s = "deeedbbcccbdaa", k = 3

Output: "aa"

Explanation:

First delete "eee" and "ccc", get "ddbbbdaa"

Then delete "bbb", get "dddaa"

Finally delete "ddd", get "aa"

Example 3:

Input: s = "pbbcggttciiippooaais", k = 2

Output: "ps"

Constraints:

• 1 < = s . l e n g t h < = 10 5 1 <= s.length <= 10^5 1<=s.length<=105
• 2 < = k < = 10 4 2 <= k <= 10^4 2<=k<=104
• s only contains lowercase English letters.

From: LeetCode

Link: 1209. Remove All Adjacent Duplicates in String II

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Solution:

Ideas:

Use a stack of pairs (char, count); whenever a count reaches k, remove it; this simulates collapsing adjacent duplicates in one pass.

Code:

#include <stdlib.h>
#include <string.h>

char* removeDuplicates(char* s, int k) {
    int n = strlen(s);

    char *stackChar = (char*)malloc(n);
    int *stackCount = (int*)malloc(n * sizeof(int));
    int top = -1;

    for (int i = 0; i < n; i++) {
        char c = s[i];

        if (top >= 0 && stackChar[top] == c) {
            stackCount[top]++;
        } else {
            top++;
            stackChar[top] = c;
            stackCount[top] = 1;
        }

        if (stackCount[top] == k) {
            top--;  // remove k duplicates
        }
    }

    char *res = (char*)malloc(n + 1);
    int idx = 0;

    for (int i = 0; i <= top; i++) {
        for (int j = 0; j < stackCount[i]; j++) {
            res[idx++] = stackChar[i];
        }
    }

    res[idx] = '\0';

    free(stackChar);
    free(stackCount);

    return res;
}