Find Target Indices After Sorting Array

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

Solutions:

C++ :

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 {
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;
    }
};

Python

def targetIndices(self, nums: List[int], target: int) -> List[int]:
    nums.sort()
    res = []
    for i, char in enumerate(nums):
        if char == target:
            res.append(i)
    return res

Comments

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