Sync LeetCode submission Runtime - 184 ms (30.91%), Memory - 15.2 MB (100.00%)

Author: jimit105

0045-jump-game-ii/README.md

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+<p>You are given a <strong>0-indexed</strong> array of integers <code>nums</code> of length <code>n</code>. You are initially positioned at <code>nums[0]</code>.</p>
+
+<p>Each element <code>nums[i]</code> represents the maximum length of a forward jump from index <code>i</code>. In other words, if you are at <code>nums[i]</code>, you can jump to any <code>nums[i + j]</code> where:</p>
+
+<ul>
+	<li><code>0 &lt;= j &lt;= nums[i]</code> and</li>
+	<li><code>i + j &lt; n</code></li>
+</ul>
+
+<p>Return <em>the minimum number of jumps to reach </em><code>nums[n - 1]</code>. The test cases are generated such that you can reach <code>nums[n - 1]</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [2,3,1,1,4]
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> 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.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [2,3,0,1,4]
+<strong>Output:</strong> 2
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
+	<li><code>0 &lt;= nums[i] &lt;= 1000</code></li>
+	<li>It&#39;s guaranteed that you can reach <code>nums[n - 1]</code>.</li>
+</ul>

0045-jump-game-ii/solution.py

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+# Approach - Greedy
+
+class Solution:
+    def jump(self, nums: List[int]) -> int:
+        jumps = 0
+        current_jump_end = 0
+        farthest = 0
+        
+        for i in range(len(nums) - 1):
+            farthest = max(farthest, i + nums[i])        
+            if i == current_jump_end:
+                jumps += 1
+                current_jump_end = farthest
+                
+        return jumps