LeetCode 1871: Jump Game VII Solution
Master LeetCode problem 1871 (Jump Game VII), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
1871. Jump Game VII
Problem Explanation
Explanation:
To solve this problem, we can use a sliding window approach along with dynamic programming. We will keep track of the valid positions we can reach at each step by maintaining a boolean array. We start from the first position and iterate through the array, updating the boolean array based on whether we can reach the current position from a valid previous position within the valid range.
At each step, we check if the current position is a '0' and if there is a valid position in the range [i - maxJump, i - minJump] from which we can reach the current position. If so, we mark the current position as reachable.
Finally, we return whether the last position is reachable or not.
Time Complexity:
The time complexity of this approach is O(n), where n is the length of the input string s.
Space Complexity:
The space complexity is O(n) for the boolean array.
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Solution Code
class Solution {
public boolean canReach(String s, int minJump, int maxJump) {
int n = s.length();
boolean[] dp = new boolean[n];
dp[0] = true;
int count = 0;
for (int i = 1; i < n; i++) {
if (s.charAt(i) == '0' && i >= minJump) {
count += dp[i - minJump] ? 1 : 0;
}
if (i > maxJump && s.charAt(i - maxJump) == '0') {
count -= dp[i - maxJump] ? 1 : 0;
}
dp[i] = count > 0;
}
return dp[n - 1];
}
}Try It Yourself
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