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Java-based LeetCode algorithm problem solutions, regularly updated
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1529\. Minimum Suffix Flips
Medium
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'`.
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