Question

1871. Jump Game VII

中文文档

Description

You are given a 0-indexed binary string s and two integers minJump and maxJump. In the beginning, you are standing at index 0, which is equal to '0'. You can move from index i to index j if the following conditions are fulfilled:

• i + minJump <= j <= min(i + maxJump, s.length - 1), and
• s[j] == '0'.

Return true if you can reach index s.length - 1 in s_, or _false_ otherwise._

 

Example 1:

Input: s = "011010", minJump = 2, maxJump = 3
Output: true
Explanation:
In the first step, move from index 0 to index 3. 
In the second step, move from index 3 to index 5.

Example 2:

Input: s = "01101110", minJump = 2, maxJump = 3
Output: false

 

Constraints:

• 2 <= s.length <= 105
• s[i] is either '0' or '1'.
• s[0] == '0'
• 1 <= minJump <= maxJump < s.length

Solutions

Solution 1: Prefix Sum + Dynamic Programming

We define a prefix sum array $pre$ of length $n+1$, where $pre[i]$ represents the number of reachable positions in the first $i$ positions of $s$. We define a boolean array $f$ of length $n$, where $f[i]$ indicates whether $s[i]$ is reachable. Initially, $pre[1] = 1$ and $f[0] = true$.

Consider $i \in [1, n)$, if $s[i] = 0$, then we need to determine whether there exists a position $j$ in the first $i$ positions of $s$, such that $j$ is reachable and the distance from $j$ to $i$ is within $[minJump, maxJump]$. If such a position $j$ exists, then we have $f[i] = true$, otherwise $f[i] = false$. When determining whether $j$ exists, we can use the prefix sum array $pre$ to determine whether such a position $j$ exists in $O(1)$ time.

The final answer is $f[n-1]$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$.

class Solution:
  def canReach(self, s: str, minJump: int, maxJump: int) -> bool:
    n = len(s)
    pre = [0] * (n + 1)
    pre[1] = 1
    f = [True] + [False] * (n - 1)
    for i in range(1, n):
      if s[i] == "0":
        l, r = max(0, i - maxJump), i - minJump
        f[i] = l <= r and pre[r + 1] - pre[l] > 0
      pre[i + 1] = pre[i] + f[i]
    return f[-1]

class Solution {
  public boolean canReach(String s, int minJump, int maxJump) {
    int n = s.length();
    int[] pre = new int[n + 1];
    pre[1] = 1;
    boolean[] f = new boolean[n];
    f[0] = true;
    for (int i = 1; i < n; ++i) {
      if (s.charAt(i) == '0') {
        int l = Math.max(0, i - maxJump);
        int r = i - minJump;
        f[i] = l <= r && pre[r + 1] - pre[l] > 0;
      }
      pre[i + 1] = pre[i] + (f[i] ? 1 : 0);
    }
    return f[n - 1];
  }
}

class Solution {
public:
  bool canReach(string s, int minJump, int maxJump) {
    int n = s.size();
    int pre[n + 1];
    memset(pre, 0, sizeof(pre));
    pre[1] = 1;
    bool f[n];
    memset(f, 0, sizeof(f));
    f[0] = true;
    for (int i = 1; i < n; ++i) {
      if (s[i] == '0') {
        int l = max(0, i - maxJump);
        int r = i - minJump;
        f[i] = l <= r && pre[r + 1] - pre[l];
      }
      pre[i + 1] = pre[i] + f[i];
    }
    return f[n - 1];
  }
};

func canReach(s string, minJump int, maxJump int) bool {
  n := len(s)
  pre := make([]int, n+1)
  pre[1] = 1
  f := make([]bool, n)
  f[0] = true
  for i := 1; i < n; i++ {
    if s[i] == '0' {
      l, r := max(0, i-maxJump), i-minJump
      f[i] = l <= r && pre[r+1]-pre[l] > 0
    }
    pre[i+1] = pre[i]
    if f[i] {
      pre[i+1]++
    }
  }
  return f[n-1]
}

function canReach(s: string, minJump: number, maxJump: number): boolean {
  const n = s.length;
  const pre: number[] = Array(n + 1).fill(0);
  pre[1] = 1;
  const f: boolean[] = Array(n).fill(false);
  f[0] = true;
  for (let i = 1; i < n; ++i) {
    if (s[i] === '0') {
      const [l, r] = [Math.max(0, i - maxJump), i - minJump];
      f[i] = l <= r && pre[r + 1] - pre[l] > 0;
    }
    pre[i + 1] = pre[i] + (f[i] ? 1 : 0);
  }
  return f[n - 1];
}

/**
 * @param {string} s
 * @param {number} minJump
 * @param {number} maxJump
 * @return {boolean}
 */
var canReach = function (s, minJump, maxJump) {
  const n = s.length;
  const pre = Array(n + 1).fill(0);
  pre[1] = 1;
  const f = Array(n).fill(false);
  f[0] = true;
  for (let i = 1; i < n; ++i) {
    if (s[i] === '0') {
      const [l, r] = [Math.max(0, i - maxJump), i - minJump];
      f[i] = l <= r && pre[r + 1] - pre[l] > 0;
    }
    pre[i + 1] = pre[i] + (f[i] ? 1 : 0);
  }
  return f[n - 1];
};