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.
Convert the previous solution to an iterative one. 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 fewest jumps required to move from the first index to the last index.
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, which can lead to incorrect solutions.
To solve this problem, we can use a greedy approach combined with dynamic programming. The idea is to keep track of the minimum number of jumps needed to reach each index. We will iterate through the array and update the minimum jumps required for each position based on the maximum jump length from previous positions.
A naive solution would involve exploring all possible paths from the first index to the last index, which would be highly inefficient and have exponential time complexity.
We can optimize the solution by using an iterative approach with dynamic programming. We maintain an array to store the minimum number of jumps required to reach each index. This approach ensures that we only compute the minimum jumps once for each index, resulting in a time complexity of O(n^2).
1. Initialize an array `jumps` of the same length as the input array, with all elements set to infinity, except the first element which is set to 0 (since no jumps are needed to reach the first index).
2. Iterate through the array using a nested loop. For each index `i`, update the minimum jumps required for all reachable indices from `i`.
3. The value at the last index of the `jumps` array will be the minimum number of jumps required to reach the last index.
#include <iostream>
#include <vector>
#include <climits> // For INT_MAX
using namespace std;
int minJumps(vector<int>& nums) {
int n = nums.size();
if (n == 0 || nums[0] == 0) return INT_MAX; // If the first element is 0, we can't move anywhere
vector<int> jumps(n, INT_MAX); // Initialize jumps array with infinity
jumps[0] = 0; // No jumps needed to reach the first index
// Iterate through the array
for (int i = 1; i < n; ++i) {
for (int j = 0; j < i; ++j) {
// If index i is reachable from index j
if (i <= j + nums[j] && jumps[j] != INT_MAX) {
jumps[i] = min(jumps[i], jumps[j] + 1);
break; // No need to check further
}
}
}
return jumps[n - 1];
}
int main() {
vector<int> nums = {2, 3, 1, 1, 4};
cout << "Minimum number of jumps: " << minJumps(nums) << endl;
return 0;
}
The time complexity of this approach is O(n^2) because of the nested loops iterating through the array. The space complexity is O(n) due to the additional `jumps` array used to store the minimum jumps required for each index.
1. If the array is empty, the function should return INT_MAX as it's not possible to reach the last index.
2. If the first element is 0, the function should return INT_MAX as it's not possible to move anywhere from the first index.
3. Arrays with only one element should return 0 as no jumps are needed.
To test the solution comprehensively, consider the following test cases:
When approaching such problems, it's essential to break down the problem into smaller subproblems and think about the constraints and requirements. 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 last index of an array. We explored a naive solution and then optimized it using an iterative approach with dynamic programming. We provided a detailed explanation of the algorithm, code implementation, complexity analysis, and testing strategies.