1871. Jump Game VII

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 \(s\).

Python3JavaC++GoTypeScriptJavaScript

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];
};

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