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];
};

1871. Jump Game VII

Description

Solutions

Solution 1: Prefix Sum + Dynamic Programming

Comments