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2311. Longest Binary Subsequence Less Than or Equal to K

Description

You are given a binary string s and a positive integer k.

Return the length of the longest subsequence of s that makes up a binary number less than or equal to k.

Note:

  • The subsequence can contain leading zeroes.
  • The empty string is considered to be equal to 0.
  • A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.

 

Example 1:

Input: s = "1001010", k = 5
Output: 5
Explanation: The longest subsequence of s that makes up a binary number less than or equal to 5 is "00010", as this number is equal to 2 in decimal.
Note that "00100" and "00101" are also possible, which are equal to 4 and 5 in decimal, respectively.
The length of this subsequence is 5, so 5 is returned.

Example 2:

Input: s = "00101001", k = 1
Output: 6
Explanation: "000001" is the longest subsequence of s that makes up a binary number less than or equal to 1, as this number is equal to 1 in decimal.
The length of this subsequence is 6, so 6 is returned.

 

Constraints:

  • 1 <= s.length <= 1000
  • s[i] is either '0' or '1'.
  • 1 <= k <= 109

Solutions

Solution 1: Greedy

The longest binary subsequence must include all the \(0\)s in the original string. On this basis, we traverse \(s\) from right to left. If we encounter a \(1\), we check if adding this \(1\) to the subsequence keeps the binary number \(v \leq k\).

The time complexity is \(O(n)\), where \(n\) is the length of the string \(s\). The space complexity is \(O(1)\).

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class Solution:
    def longestSubsequence(self, s: str, k: int) -> int:
        ans = v = 0
        for c in s[::-1]:
            if c == "0":
                ans += 1
            elif ans < 30 and (v | 1 << ans) <= k:
                v |= 1 << ans
                ans += 1
        return ans
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class Solution {
    public int longestSubsequence(String s, int k) {
        int ans = 0, v = 0;
        for (int i = s.length() - 1; i >= 0; --i) {
            if (s.charAt(i) == '0') {
                ++ans;
            } else if (ans < 30 && (v | 1 << ans) <= k) {
                v |= 1 << ans;
                ++ans;
            }
        }
        return ans;
    }
}
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class Solution {
public:
    int longestSubsequence(string s, int k) {
        int ans = 0, v = 0;
        for (int i = s.size() - 1; ~i; --i) {
            if (s[i] == '0') {
                ++ans;
            } else if (ans < 30 && (v | 1 << ans) <= k) {
                v |= 1 << ans;
                ++ans;
            }
        }
        return ans;
    }
};
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func longestSubsequence(s string, k int) (ans int) {
    for i, v := len(s)-1, 0; i >= 0; i-- {
        if s[i] == '0' {
            ans++
        } else if ans < 30 && (v|1<<ans) <= k {
            v |= 1 << ans
            ans++
        }
    }
    return
}
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function longestSubsequence(s: string, k: number): number {
    let ans = 0;
    for (let i = s.length - 1, v = 0; ~i; --i) {
        if (s[i] == '0') {
            ++ans;
        } else if (ans < 30 && (v | (1 << ans)) <= k) {
            v |= 1 << ans;
            ++ans;
        }
    }
    return ans;
}
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/**
 * @param {string} s
 * @param {number} k
 * @return {number}
 */
var longestSubsequence = function (s, k) {
    let ans = 0;
    for (let i = s.length - 1, v = 0; ~i; --i) {
        if (s[i] == '0') {
            ++ans;
        } else if (ans < 30 && (v | (1 << ans)) <= k) {
            v |= 1 << ans;
            ++ans;
        }
    }
    return ans;
};
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public class Solution {
    public int LongestSubsequence(string s, int k) {
        int ans = 0, v = 0;
        for (int i = s.Length - 1; i >= 0; --i) {
            if (s[i] == '0') {
                ++ans;
            } else if (ans < 30 && (v | 1 << ans) <= k) {
                v |= 1 << ans;
                ++ans;
            }
        }
        return ans;
    }
}

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