g1701_1800.s1702_maximum_binary_string_after_change.Solution Maven / Gradle / Ivy

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package g1701_1800.s1702_maximum_binary_string_after_change;

// #Medium #String #Greedy #2022_04_24_Time_42_ms_(82.86%)_Space_80.4_MB_(50.00%)

/**
 * 1702 - Maximum Binary String After Change\.
 *
 * Medium
 *
 * You are given a binary string `binary` consisting of only `0`'s or `1`'s. You can apply each of the following operations any number of times:
 *
 * *   Operation 1: If the number contains the substring `"00"`, you can replace it with `"10"`.
 *     *   For example, `"00010" -> "10010`"
 * *   Operation 2: If the number contains the substring `"10"`, you can replace it with `"01"`.
 *     *   For example, `"00010" -> "00001"`
 *
 * _Return the **maximum binary string** you can obtain after any number of operations. Binary string `x` is greater than binary string `y` if `x`'s decimal representation is greater than `y`'s decimal representation._
 *
 * **Example 1:**
 *
 * **Input:** binary = "000110"
 *
 * **Output:** "111011"
 *
 * **Explanation:** A valid transformation sequence can be: 
 *
 * "000110" -> "000101"
 *
 * "000101" -> "100101" 
 *
 * "100101" -> "110101" 
 *
 * "110101" -> "110011" 
 *
 * "110011" -> "111011"
 *
 * **Example 2:**
 *
 * **Input:** binary = "01"
 *
 * **Output:** "01"
 *
 * **Explanation:** "01" cannot be transformed any further.
 *
 * **Constraints:**
 *
 * *   1 <= binary.length <= 10⁵
 * *   `binary` consist of `'0'` and `'1'`.
**/
public class Solution {
    public String maximumBinaryString(String binary) {
        char[] bs = binary.toCharArray();
        int zcount = 0;
        int pos = -1;
        for (int i = bs.length - 1; i >= 0; i--) {
            if (bs[i] == '0') {
                bs[i] = '1';
                zcount++;
                pos = i;
            }
        }
        if (pos >= 0) {
            bs[pos + zcount - 1] = '0';
        }
        return new String(bs);
    }
}