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Formatted question description: https://leetcode.ca/all/1920.html

1920. Build Array from Permutation

Level

Easy

Description

Given a zero-based permutation nums (0-indexed), build an array ans of the same length where ans[i] = nums[nums[i]] for each 0 <= i < nums.length and return it.

A zero-based permutation nums is an array of distinct integers from 0 to nums.length - 1 (inclusive).

Example 1:

Input: nums = [0,2,1,5,3,4]
Output: [0,1,2,4,5,3]
Explanation: The array ans is built as follows: 
ans = [nums[nums[0]], nums[nums[1]], nums[nums[2]], nums[nums[3]], nums[nums[4]], nums[nums[5]]]
    = [nums[0], nums[2], nums[1], nums[5], nums[3], nums[4]]
    = [0,1,2,4,5,3]
**Example 2:**

Input: nums = [5,0,1,2,3,4] Output: [4,5,0,1,2,3] Explanation: The array ans is built as follows: ans = [nums[nums[0]], nums[nums[1]], nums[nums[2]], nums[nums[3]], nums[nums[4]], nums[nums[5]]] = [nums[5], nums[0], nums[1], nums[2], nums[3], nums[4]] = [4,5,0,1,2,3] ```

Constraints:

• 1 <= nums.length <= 1000
• 0 <= nums[i] < nums.length
• The elements in nums are distinct.

Solution

Simply create the array ans as described in the problem, and return ans.

• Java
• C++
• Python
• Go
• TypeScript
• Javascript
• RenderScript

• class Solution {
      public int[] buildArray(int[] nums) {
          int length = nums.length;
          int[] array = new int[length];
          for (int i = 0; i < length; i++)
              array[i] = nums[nums[i]];
          return array;
      }
  }
  
  ############
  
  class Solution {
      public int[] buildArray(int[] nums) {
          int[] ans = new int[nums.length];
          for (int i = 0; i < nums.length; ++i) {
              ans[i] = nums[nums[i]];
          }
          return ans;
      }
  }
  

• // OJ: https://leetcode.com/problems/build-array-from-permutation/
  // Time: O(N)
  // Space: O(1)
  class Solution {
  public:
      vector<int> buildArray(vector<int>& A) {
          vector<int> ans(A.size());
          for (int i = 0; i < A.size(); ++i) {
              ans[i] = A[A[i]];
          }
          return ans;
      }
  };
  

• class Solution:
      def buildArray(self, nums: List[int]) -> List[int]:
          return [nums[num] for num in nums]
  
  ############
  
  # 1920. Build Array from Permutation
  # https://leetcode.com/problems/build-array-from-permutation/
  
  class Solution:
      def buildArray(self, nums: List[int]) -> List[int]:
          res = [0] * len(nums)
          
          for i in range(len(nums)):
              res[i] = nums[nums[i]]
          
          return res
  
  

• func buildArray(nums []int) []int {
  	ans := make([]int, len(nums))
  	for i, num := range nums {
  		ans[i] = nums[num]
  	}
  	return ans
  }
  

• function buildArray(nums: number[]): number[] {
      return nums.map(v => nums[v]);
  }
  
  

• /**
   * @param {number[]} nums
   * @return {number[]}
   */
  var buildArray = function (nums) {
      let ans = [];
      for (let i = 0; i < nums.length; ++i) {
          ans[i] = nums[nums[i]];
      }
      return ans;
  };
  
  

• impl Solution {
      pub fn build_array(nums: Vec<i32>) -> Vec<i32> {
          nums.iter().map(|&v| nums[v as usize]).collect()
      }
  }
  
  

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