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Formatted question description: https://leetcode.ca/all/2089.html

2089. Find Target Indices After Sorting Array (Easy)

You are given a 0-indexed integer array nums and a target element target.

A target index is an index i such that nums[i] == target.

Return a list of the target indices of nums after sorting nums in non-decreasing order. If there are no target indices, return an empty list. The returned list must be sorted in increasing order.

 

Example 1:

Input: nums = [1,2,5,2,3], target = 2
Output: [1,2]
Explanation: After sorting, nums is [1,2,2,3,5].
The indices where nums[i] == 2 are 1 and 2.

Example 2:

Input: nums = [1,2,5,2,3], target = 3
Output: [3]
Explanation: After sorting, nums is [1,2,2,3,5].
The index where nums[i] == 3 is 3.

Example 3:

Input: nums = [1,2,5,2,3], target = 5
Output: [4]
Explanation: After sorting, nums is [1,2,2,3,5].
The index where nums[i] == 5 is 4.

Example 4:

Input: nums = [1,2,5,2,3], target = 4
Output: []
Explanation: There are no elements in nums with value 4.

 

Constraints:

• 1 <= nums.length <= 100
• 1 <= nums[i], target <= 100

Similar Questions:

• Find First and Last Position of Element in Sorted Array (Medium)
• Rank Transform of an Array (Easy)

Solution 1. Sorting

• C++
• Python
• Java
• Go
• TypeScript

• // OJ: https://leetcode.com/problems/find-target-indices-after-sorting-array/
  // Time: O(NlogN)
  // Space: O(1) extra space
  class Solution {
  public:
      vector<int> targetIndices(vector<int>& A, int target) {
          sort(begin(A), end(A));
          vector<int> ans;
          for (int i = 0; i < A.size(); ++i) {
              if (A[i] == target) ans.push_back(i);
          }
          return ans;
      }
  };
  

• class Solution:
      def targetIndices(self, nums: List[int], target: int) -> List[int]:
          nums.sort()
          return [i for i, v in enumerate(nums) if v == target]
  
  ############
  
  # 2089. Find Target Indices After Sorting Array
  # https://leetcode.com/problems/find-target-indices-after-sorting-array/
  
  class Solution:
      def targetIndices(self, nums: List[int], target: int) -> List[int]:
          nums.sort()
          res = []
          
          for i, x in enumerate(nums):
              if x == target:
                  res.append(i)
          
          return res
  
  

• class Solution {
      public List<Integer> targetIndices(int[] nums, int target) {
          Arrays.sort(nums);
          List<Integer> ans = new ArrayList<>();
          for (int i = 0; i < nums.length; ++i) {
              if (nums[i] == target) {
                  ans.add(i);
              }
          }
          return ans;
      }
  }
  

• func targetIndices(nums []int, target int) (ans []int) {
  	sort.Ints(nums)
  	for i, v := range nums {
  		if v == target {
  			ans = append(ans, i)
  		}
  	}
  	return
  }
  

• function targetIndices(nums: number[], target: number): number[] {
      nums.sort((a, b) => a - b);
      let ans: number[] = [];
      for (let i = 0; i < nums.length; ++i) {
          if (nums[i] == target) {
              ans.push(i);
          }
      }
      return ans;
  }
  
  

Solution 2. Counting Sort

• C++
• Python
• Java
• Go
• TypeScript

• // OJ: https://leetcode.com/problems/find-target-indices-after-sorting-array/
  // Time: O(N)
  // Space: O(1)
  class Solution {
  public:
      vector<int> targetIndices(vector<int>& A, int target) {
          int cnt = 0, rank = 0; // `cnt` is the frequency of `target`, `rank` is the sum of frequencies of all numbers < target
          for (int n : A) {
              cnt += n == target;
              rank += n < target;
          }
          vector<int> ans;
          while (cnt--) ans.push_back(rank++);
          return ans;
      }
  };
  

• class Solution:
      def targetIndices(self, nums: List[int], target: int) -> List[int]:
          nums.sort()
          return [i for i, v in enumerate(nums) if v == target]
  
  ############
  
  # 2089. Find Target Indices After Sorting Array
  # https://leetcode.com/problems/find-target-indices-after-sorting-array/
  
  class Solution:
      def targetIndices(self, nums: List[int], target: int) -> List[int]:
          nums.sort()
          res = []
          
          for i, x in enumerate(nums):
              if x == target:
                  res.append(i)
          
          return res
  
  

• class Solution {
      public List<Integer> targetIndices(int[] nums, int target) {
          Arrays.sort(nums);
          List<Integer> ans = new ArrayList<>();
          for (int i = 0; i < nums.length; ++i) {
              if (nums[i] == target) {
                  ans.add(i);
              }
          }
          return ans;
      }
  }
  

• func targetIndices(nums []int, target int) (ans []int) {
  	sort.Ints(nums)
  	for i, v := range nums {
  		if v == target {
  			ans = append(ans, i)
  		}
  	}
  	return
  }
  

• function targetIndices(nums: number[], target: number): number[] {
      nums.sort((a, b) => a - b);
      let ans: number[] = [];
      for (let i = 0; i < nums.length; ++i) {
          if (nums[i] == target) {
              ans.push(i);
          }
      }
      return ans;
  }
  
  

Discuss

https://leetcode.com/problems/find-target-indices-after-sorting-array/discuss/1599800/C%2B%2B-O(N)-Time-Counting-Sort

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