Merge pull request #1136 from PradnyaGaitonde/PradnyaGaitonde-patch-4

Create 0045-jump-game-II.md

Author: ajay-dhangar

dsa-solutions/lc-solutions/0000-0099/0045-jump-game-II.md

@@ -0,0 +1,153 @@
+---
+id: jump-game-II
+title: Jump Game II(LeetCode)
+sidebar_label: 0045-Jump-Game-II
+tags:
+  - Array
+  - Dynamic Programming
+  - Greedy
+description: You are given a 0-indexed array of integers nums of length `n`. Return the minimum number of jumps to reach `nums[n - 1]`.
+sidebar_position: 45
+---
+
+## Problem Statement
+
+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]` and
+- `i + j < n`
+Return the minimum number of jumps to reach `nums[n - 1]`. The test cases are generated such that you can reach `nums[n - 1]`.
+
+### Examples
+
+**Example 1:**
+
+```plaintext
+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:**
+
+```plaintext
+Input: nums = [2,3,0,1,4]
+Output: 2
+```
+
+### Constraints
+
+- `1 <= nums.length <= 104`
+- `0 <= nums[i] <= 1000`
+- It's guaranteed that you can reach `nums[n - 1]`.
+
+## Solution
+
+In this problem, we aim to find the minimum number of jumps required to reach the end of the array, starting from the first element. We explore three main approaches: Recursive Dynamic Programming with 
+Memoization, Iterative Dynamic Programming with Tabulation, and a Greedy BFS approach.
+
+### Approach 1: Recursive Dynamic Programming (Memoization)
+Concept: Store the solutions for each position to avoid redundant calculations.
+
+#### Algorithm
+
+1. Initialize `dp` array with a large value to indicate uncomputed positions.
+2. Call the recursive function `solve` starting from position 0.
+3. In `solve`:
+* If the position is at or beyond the end, return 0.
+* If `dp[pos]` is already computed, return its value.
+* Explore all possible jumps from the current position.
+* Store and return the minimum jumps required.
+
+#### Implementation
+
+```C++
+int jump(vector<int>& nums) {
+    vector<int> dp(size(nums), 10001); 
+    return solve(nums, dp, 0);
+}
+
+int solve(vector<int>& nums, vector<int>& dp, int pos) {
+    if(pos >= size(nums) - 1) return 0; 
+    if(dp[pos] != 10001) return dp[pos];
+    for(int j = 1; j <= nums[pos]; j++)
+        dp[pos] = min(dp[pos], 1 + solve(nums, dp, pos + j));        
+    return dp[pos];
+}
+```
+
+### Complexity Analysis
+
+- **Time complexity**: O(N^2)
+- **Space complexity**: O(N)
+
+### Approach 2: Iterative Dynamic Programming (Tabulation)
+
+Concept: Start from the last index and iteratively compute the minimum jumps required for each position.
+
+#### Algorithm
+
+1. Initialize `dp` array with large values and set `dp[n-1]` to 0.
+2. Iterate from the second last index to the start.
+3. For each index, explore all possible jumps and store the minimum jumps required.
+   
+#### Implementation 
+
+```C++
+int jump(vector<int>& nums) {
+    int n = size(nums);
+    vector<int> dp(n, 10001);
+    dp[n - 1] = 0;  
+    for(int i = n - 2; i >= 0; i--) 
+        for(int jumpLen = 1; jumpLen <= nums[i]; jumpLen++) 
+            dp[i] = min(dp[i], 1 + dp[min(n - 1, i + jumpLen)]);
+    return dp[0];
+}
+```
+
+### Complexity Analysis
+
+- **Time complexity**: O(N^2)
+- **Space complexity**: O(N)
+
+### Approach 3: Greedy BFS
+
+Concept: Use two pointers to track the furthest reachable position 
+and the currently furthest reached position. Update the jump count when moving to a new level.
+
+#### Algorithm
+
+1. Initialize `maxReachable`, `lastJumpedPos`, and `jumps`.
+2. Iterate over the array until the lastJumpedPos reaches or exceeds the last index.
+3. Update `maxReachable` with the furthest position reachable from the current index.
+4. When `i` equals `lastJumpedPos`, move to `maxReachable` and increment jumps.
+   
+#### Implementation 
+
+```C++
+int jump(vector<int>& nums) {
+    int n = size(nums), i = 0, maxReachable = 0, lastJumpedPos = 0, jumps = 0;
+    while(lastJumpedPos < n - 1) {
+        maxReachable = max(maxReachable, i + nums[i]);
+        if(i == lastJumpedPos) {
+            lastJumpedPos = maxReachable;
+            jumps++;
+        }
+        i++;
+    }
+    return jumps;
+}
+```
+
+### Complexity Analysis
+
+- **Time complexity**: O(N)
+- **Space complexity**: O(1)
+
+### Conclusion
+
+1. Recursive DP with Memoization: Efficiently avoids redundant calculations but has higher time complexity.
+2. Iterative DP with Tabulation: Iteratively solves for each position but has quadratic time complexity.
+3. Greedy BFS: Provides the most optimal solution with linear time complexity and constant space, making it the best approach for this problem.