2599. Take K Of Each Character From Left And Right¶

Difficulty: Medium

2599. Take K of Each Character From Left and Right

Medium

You are given a string s consisting of the characters 'a', 'b', and 'c' and a non-negative integer k. Each minute, you may take either the leftmost character of s, or the rightmost character of s.

Return the minimum number of minutes needed for you to take at least k of each character, or return -1 if it is not possible to take k of each character.

Example 1:

Input: s = "aabaaaacaabc", k = 2
Output: 8
Explanation: 
Take three characters from the left of s. You now have two 'a' characters, and one 'b' character.
Take five characters from the right of s. You now have four 'a' characters, two 'b' characters, and two 'c' characters.
A total of 3 + 5 = 8 minutes is needed.
It can be proven that 8 is the minimum number of minutes needed.

Example 2:

Input: s = "a", k = 1
Output: -1
Explanation: It is not possible to take one 'b' or 'c' so return -1.

Constraints:

1 <= s.length <= 10⁵
s consists of only the letters 'a', 'b', and 'c'.
0 <= k <= s.length

Solution¶

class Solution {
    private int pref[][];
    private int suff[][];
    public int takeCharacters(String s, int k) {
        int n = s.length();
        if (k == 0) return 0;
        pref = new int[n][3];
        suff = new int[n][3];
        int counta = 0, countb = 0, countc = 0;
        for (int i = 0; i < n; i++) {
            if (s.charAt(i) == 'a') counta++;
            else if (s.charAt(i) == 'b') countb++;
            else countc++;
            pref[i][0] = counta; pref[i][1] = countb; pref[i][2] = countc;
        }
        counta = 0; countb = 0; countc = 0;
        for (int i = n - 1; i >= 0; i--) {
            if (s.charAt(i) == 'a') counta++;
            else if (s.charAt(i) == 'b') countb++;
            else countc++;
            suff[i][0] = counta; suff[i][1] = countb; suff[i][2] = countc;
        }
        if (pref[n - 1][0] < k || pref[n - 1][1] < k || pref[n - 1][2] < k) return -1; 
        int mini = Integer.MAX_VALUE;
        for (int i = 0; i < n; i++) {
            int curr_counta = pref[i][0], curr_countb = pref[i][1], curr_countc = pref[i][2];
            if (curr_counta >= k && curr_countb >= k && curr_countc >= k) {
                mini = Math.min(mini, i + 1);
                continue;
            }
            int ind = binary_search(i + 1, n - 1, n, k);
            if (ind == -1) continue;
            mini = Math.min(mini, (i + 1 + (n  - ind)));
        }

        for (int i = n - 1; i >= 0; i--) {
            int curr_counta = suff[i][0], curr_countb = suff[i][1], curr_countc = suff[i][2];
            if (curr_counta >= k && curr_countb >= k && curr_countc >= k) {
                mini = Math.min(mini, (n - i));
                continue;
            }
            if (i - 1 >= 0) {
                int ind = binary_search1(0, i - 1, n , k);
                if (ind != -1) mini = Math.min(mini, (n - i + 1 + ind + 1)); 
            }
        }
        return mini;
    }
    private int binary_search(int low, int high, int n, int k) {
        int counta = 0, countb = 0, countc = 0;
        counta = pref[low - 1][0]; countb = pref[low - 1][1]; countc = pref[low - 1][2];
        int ans = -1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            int curr_counta = suff[mid][0], curr_countb = suff[mid][1], curr_countc = suff[mid][2];
            if (counta + curr_counta >= k && countb + curr_countb >= k && countc + curr_countc >= k) {
                ans = mid;
                low = mid + 1;
            } 
            else high = mid - 1;
        }
        return ans;
    }

    private int binary_search1(int low, int high, int n, int k) {
        int counta = suff[high + 1][0], countb = suff[high + 1][1], countc = suff[high + 1][2];
        int ans = -1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            int curr_counta = pref[mid][0], curr_countb = pref[mid][1], curr_countc = pref[mid][2];
            if (counta + curr_counta >= k && countb + curr_countb >= k && countc + curr_countc >= k) {
                ans = mid;
                high = mid - 1;
            }
            else low = mid + 1;
        }
        return ans;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here