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Formatted question description: https://leetcode.ca/all/2311.html

# 2311. Longest Binary Subsequence Less Than or Equal to K

• Difficulty: Medium.
• Related Topics: String, Dynamic Programming, Greedy, Memoization.
• Similar Questions: Maximum Binary String After Change.

## Problem

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

## Solution (Java, C++, Python)

• class Solution {
public int longestSubsequence(String s, int k) {
int res = 0;
int cost = 1;
int n = s.length();
for (int i = n - 1; i >= 0; i--) {
if (s.charAt(i) == '0' || cost <= k) {
k -= cost * (s.charAt(i) - '0');
++res;
}
if (cost <= k) {
cost *= 2;
}
}
return res;
}
}

############

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;
}
}

• 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

############

# 2311. Longest Binary Subsequence Less Than or Equal to K
# https://leetcode.com/problems/longest-binary-subsequence-less-than-or-equal-to-k

class Solution:
def longestSubsequence(self, s: str, k: int) -> int:
n = len(s)
res = s.count("0")
val = 0
base = 1

for x in s[::-1]:
if val + base > k: break

if x == "1":
res += 1
val += base

base *= 2

return res


• 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;
}
};

• 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
}

• 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;
}


• /**
* @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;
};



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).