You are given a 0-indexed array of integers nums of length n. You are initially positioned at nums[0].
Each element nums[i] represents the maximum length of a forward jump from index i. In other words, if you are at nums[i], you can jump to any nums[i + j] where:
0 <= j <= nums[i] andi + j < nReturn the minimum number of jumps to reach nums[n - 1]. The test cases are generated such that you can reach nums[n - 1].
Example 1:
Input: nums = [2,3,1,1,4] Output: 2 Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.
Example 2:
Input: nums = [2,3,0,1,4] Output: 2
Constraints:
1 <= nums.length <= 1040 <= nums[i] <= 1000nums[n - 1].The goal of Jump Game II is to determine the minimum number of jumps needed to reach the last index of an array, where each element represents the maximum jump length from that position. A naive way to think about the problem is using Dynamic Programming. For each index, you calculate the minimum jumps required by checking all previous positions that could reach it. While intuitive, this approach can lead to O(n^2) time complexity and may be inefficient for large inputs.
A more optimal strategy uses a Greedy approach. Instead of computing jumps for every position, you track the current range of reachable indices and continuously expand the farthest position you can reach while scanning the array. Each time you exhaust the current range, you increment the jump count and update the boundary. This technique effectively simulates level-by-level traversal similar to BFS on ranges.
The greedy method achieves O(n) time complexity with O(1) extra space, making it the preferred solution for interview settings.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Dynamic Programming | O(n^2) | O(n) |
| Greedy Range Expansion | O(n) | O(1) |
NeetCode
The greedy approach involves making a jump only when absolutely necessary. Track the maximum index you can reach at each step and make a decision to jump when you get to the maximum of the current window. This ensures the least number of jumps.
Time Complexity: O(n), where n is the number of elements in the array, because we make a single pass through the array.
Space Complexity: O(1), since we are using a fixed amount of extra space.
1#include <vector>
2#include <iostream>
3using namespace std;
4
5int jump(vector<int>& nums) {
6 int jumps = 0, currentEnd = 0, farthest = 0;
7 for (int i = 0; i < nums.size() - 1; ++i) {
8 farthest = max(farthest, i + nums[i]);
9 if (i == currentEnd) {
10 jumps++;
11 currentEnd = farthest;
12 }
13 }
14 return jumps;
}
int main() {
vector<int> nums = {2, 3, 1, 1, 4};
cout << "Minimum jumps: " << jump(nums) << endl;
return 0;
}The algorithm keeps track of the maximum distance that can be reached `farthest` and increments the jump count when reaching the current jump's end. Update the end to the farthest position reached so far.
The dynamic programming approach calculates the minimum jumps required to reach each index. For each index, calculate the minimum number of jumps required from all previous indices that can reach the current index. However, this approach is less efficient in terms of time complexity.
Time Complexity: O(n^2), where n is the number of elements.
Space Complexity: O(n), due to the use of a DP array.
1
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Yes, Jump Game II is a popular interview problem because it tests greedy thinking, range tracking, and optimization skills. Variants of this problem are commonly asked in technical interviews at companies like Amazon, Google, and Meta.
The optimal approach uses a greedy strategy that tracks the current reachable range and the farthest position reachable within that range. When the current range ends, you increase the jump count and update the range to the farthest reachable index. This results in O(n) time complexity and constant extra space.
The greedy solution does not require any special data structure beyond simple variables to track boundaries and the farthest reachable index. This keeps the implementation efficient with O(1) additional space.
Dynamic programming checks many possible transitions for each index, which can lead to O(n^2) time complexity. The greedy approach avoids redundant calculations by expanding the farthest reachable boundary while scanning the array once.
The DP array `dp` is initialized with infinity. Iterate through the array to populate each index with the minimum number of jumps needed to reach it.