LeetCode 1340: Jump Game V Solution
Master LeetCode problem 1340 (Jump Game V), a hard 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.
1340. Jump Game V
Problem Explanation
Explanation:
To solve this problem, we can use dynamic programming with memoization. We will start at each index of the array and perform a recursive function to calculate the maximum number of indices we can visit from that index. The base case of the recursion will be when we have already calculated the maximum number of indices we can visit from a particular index. We will keep track of this information in a memoization array.
At each index, we will check all possible jumps within the range of d and satisfying the condition arr[i] > arr[j] to see how far we can jump. We will recursively call the function for each reachable index and take the maximum of all such values.
The time complexity of this approach is O(n * d) and the space complexity is O(n) where n is the length of the input array.
:
Solution Code
class Solution {
public int maxJumps(int[] arr, int d) {
int n = arr.length;
int[] memo = new int[n];
int maxIndices = 0;
for (int i = 0; i < n; i++) {
maxIndices = Math.max(maxIndices, dfs(arr, d, i, memo));
}
return maxIndices;
}
private int dfs(int[] arr, int d, int index, int[] memo) {
if (memo[index] != 0) {
return memo[index];
}
int maxIndices = 1; // At least one index is reachable
for (int dir = -1; dir <= 1; dir += 2) {
for (int i = 1; i <= d; i++) {
int nextIndex = index + dir * i;
if (nextIndex < 0 || nextIndex >= arr.length || arr[nextIndex] >= arr[index]) {
break;
}
maxIndices = Math.max(maxIndices, 1 + dfs(arr, d, nextIndex, memo));
}
}
memo[index] = maxIndices;
return maxIndices;
}
}Try It Yourself
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