Jump Game VII

// Accepted solution for LeetCode #1871: Jump Game VII
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];
    }
}

// Accepted solution for LeetCode #1871: Jump Game VII
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]
}

# Accepted solution for LeetCode #1871: Jump Game VII
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]

// Accepted solution for LeetCode #1871: Jump Game VII
/**
 * [1871] Jump Game VII
 *
 * 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 = "<u>0</u>11<u>0</u>1<u>0</u>", 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 <= 10^5
 * 	s[i] is either '0' or '1'.
 * 	s[0] == '0'
 * 	1 <= minJump <= maxJump < s.length
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/jump-game-vii/
// discuss: https://leetcode.com/problems/jump-game-vii/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn can_reach(s: String, min_jump: i32, max_jump: i32) -> bool {
        false
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_1871_example_1() {
        let s = "011010".to_string();
        let min_jump = 2;
        let max_jump = 3;

        let result = true;

        assert_eq!(Solution::can_reach(s, min_jump, max_jump), result);
    }

    #[test]
    #[ignore]
    fn test_1871_example_2() {
        let s = "01101110".to_string();
        let min_jump = 2;
        let max_jump = 3;

        let result = false;

        assert_eq!(Solution::can_reach(s, min_jump, max_jump), result);
    }
}

// Accepted solution for LeetCode #1871: Jump Game VII
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];
}