Given an array of non-negative integers, you are initially positioned at the first index of the array.
Each element in the array represents your maximum jump length at that position.
Your goal is to reach the last index in the minimum number of jumps.
Example:
Input: [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.
Apply memoization to the previous solution. Your algorithm should run in O(n^2) time and use O(n) extra space.
The core challenge of this problem is to find the minimum number of jumps needed to reach the last index of the array. Each element in the array represents the maximum number of steps you can jump forward from that position. The goal is to determine the minimum jumps required to reach the end of the array.
This problem is significant in scenarios where you need to find the shortest path or minimum steps in a constrained environment. Common applications include game development, pathfinding algorithms, and network routing.
Potential pitfalls include misunderstanding the jump lengths and not considering all possible paths, leading to suboptimal solutions.
To solve this problem, we can use a dynamic programming approach with memoization to store the results of subproblems and avoid redundant calculations.
A naive solution would involve exploring all possible paths from the start to the end and counting the jumps. This approach is not optimal as it has exponential time complexity.
We can optimize the solution using dynamic programming with memoization. The idea is to use an array to store the minimum number of jumps needed to reach each index. We iterate through the array and update the minimum jumps for each position based on the maximum jump length at that position.
1. Initialize an array `jumps` where `jumps[i]` represents the minimum number of jumps needed to reach index `i`.
2. Set `jumps[0]` to 0 as no jumps are needed to reach the first index.
3. Iterate through the array and for each position, update the minimum jumps for the reachable positions based on the maximum jump length at that position.
4. Return the value at the last index of the `jumps` array, which represents the minimum number of jumps needed to reach the end.
#include <vector>
#include <algorithm>
#include <climits>
using namespace std;
// Function to find the minimum number of jumps to reach the end
int jump(vector<int>& nums) {
int n = nums.size();
if (n == 0 || n == 1) return 0;
// Initialize jumps array with maximum integer values
vector<int> jumps(n, INT_MAX);
jumps[0] = 0;
// Iterate through the array
for (int i = 0; i < n; ++i) {
// Update the minimum jumps for reachable positions
for (int j = i + 1; j <= i + nums[i] && j < n; ++j) {
jumps[j] = min(jumps[j], jumps[i] + 1);
}
}
// Return the minimum jumps needed to reach the last index
return jumps[n - 1];
}
The time complexity of this approach is O(n^2) because we have nested loops iterating through the array. The space complexity is O(n) due to the additional `jumps` array used for memoization.
Potential edge cases include:
To test the solution comprehensively, consider the following test cases:
When approaching such problems, consider breaking down the problem into smaller subproblems and using dynamic programming to store intermediate results. Practice solving similar problems and study different algorithms to improve problem-solving skills.
In this blog post, we discussed the problem of finding the minimum number of jumps to reach the end of an array. We explored a dynamic programming approach with memoization to optimize the solution. Understanding and solving such problems is crucial for developing efficient algorithms and improving problem-solving skills.
For further reading and practice, consider the following resources: