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 <= 1001 <= nums[i], target <= 100
Similar Questions:
// OJ: https://leetcode.com/problems/find-target-indices-after-sorting-array/
// Author: github.com/lzl124631x
// 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;
}
};// OJ: https://leetcode.com/problems/find-target-indices-after-sorting-array/
// Author: github.com/lzl124631x
// 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;
}
};