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1529. Minimum Suffix Flips

Description

You are given a 0-indexed binary string target of length n. You have another binary string s of length n that is initially set to all zeros. You want to make s equal to target.

In one operation, you can pick an index i where 0 <= i < n and flip all bits in the inclusive range [i, n - 1]. Flip means changing '0' to '1' and '1' to '0'.

Return the minimum number of operations needed to make s equal to target.

 

Example 1:

Input: target = "10111"
Output: 3
Explanation: Initially, s = "00000".
Choose index i = 2: "00000" -> "00111"
Choose index i = 0: "00111" -> "11000"
Choose index i = 1: "11000" -> "10111"
We need at least 3 flip operations to form target.

Example 2:

Input: target = "101"
Output: 3
Explanation: Initially, s = "000".
Choose index i = 0: "000" -> "111"
Choose index i = 1: "111" -> "100"
Choose index i = 2: "100" -> "101"
We need at least 3 flip operations to form target.

Example 3:

Input: target = "00000"
Output: 0
Explanation: We do not need any operations since the initial s already equals target.

 

Constraints:

• n == target.length
• 1 <= n <= 105
• target[i] is either '0' or '1'.

Solutions

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• class Solution {
      public int minFlips(String target) {
          int ans = 0;
          for (int i = 0; i < target.length(); ++i) {
              int v = target.charAt(i) - '0';
              if (((ans & 1) ^ v) != 0) {
                  ++ans;
              }
          }
          return ans;
      }
  }
  

• class Solution {
  public:
      int minFlips(string target) {
          int ans = 0;
          for (char c : target) {
              int v = c - '0';
              if ((ans & 1) ^ v) {
                  ++ans;
              }
          }
          return ans;
      }
  };
  

• class Solution:
      def minFlips(self, target: str) -> int:
          ans = 0
          for v in target:
              if (ans & 1) ^ int(v):
                  ans += 1
          return ans
  
  

• func minFlips(target string) int {
  	ans := 0
  	for _, c := range target {
  		v := int(c - '0')
  		if ((ans & 1) ^ v) != 0 {
  			ans++
  		}
  	}
  	return ans
  }
  

• function minFlips(target: string): number {
      let ans = 0;
      for (const c of target) {
          if (ans % 2 !== +c) {
              ++ans;
          }
      }
      return ans;
  }
  
  

• impl Solution {
      pub fn min_flips(target: String) -> i32 {
          let mut ans = 0;
          for c in target.chars() {
              let bit = (c as u8 - b'0') as i32;
              if ans % 2 != bit {
                  ans += 1;
              }
          }
          ans
      }
  }
  
  

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