LeetCode 2770: Maximum Number of Jumps to Reach the Last Index Solution
Master LeetCode problem 2770 (Maximum Number of Jumps to Reach the Last Index), 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.
2770. Maximum Number of Jumps to Reach the Last Index
Problem Explanation
Explanation
To solve this problem, we can use a dynamic programming approach. We will maintain an array dp where dp[i] represents the maximum number of jumps to reach index i. We initialize dp[0] to 1 as we start at index 0. Then, for each index i, we iterate over all previous indices j where the difference between nums[j] and nums[i] is within the target range. We update dp[i] to the maximum of dp[j] + 1 for all valid j. Finally, the answer will be the maximum value in the dp array.
The time complexity of this approach is O(n^2) where n is the length of the input array nums. The space complexity is O(n).
Solution Code
class Solution {
public int maxJumps(int[] nums, int target) {
int n = nums.length;
int[] dp = new int[n];
dp[0] = 1;
for (int i = 1; i < n; i++) {
dp[i] = 1;
for (int j = i - 1; j >= 0 && i - j <= target && nums[i] - nums[j] <= target; j--) {
dp[i] = Math.max(dp[i], dp[j] + 1);
}
}
int maxJumps = 0;
for (int jumps : dp) {
maxJumps = Math.max(maxJumps, jumps);
}
return maxJumps;
}
}Try It Yourself
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