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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: 2Explanation: 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] <= 1000It's guaranteed that you can reach nums[n - 1].

Question

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: 2Explanation: 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] <= 1000It's guaranteed that you can reach nums[n - 1].

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Solution

To solve this problem, we can use a greedy algorithm. Here are the steps:

  1. Initialize two variables, jumps and farthest, to 0. jumps will keep track of the minimum number of jumps needed to reach the end, and farthest will keep track of the farthest index we can reach with the current number of jumps.

  2. Iterate over the array from the start to the second last element. For each element, update farthest to be the maximum of farthest and i + nums[i], where i is the current index.

  3. Initialize a variable end to 0. end will keep track of the end of the current jump. If we reach end, it means we have to make another jump to proceed further. So, when i equals end, increment jumps by 1 and update end to farthest.

  4. Continue this process until we reach the end of the array.

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This problem has been solved

Similar Questions

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