Given the array nums
, for each nums[i]
find out how many numbers in the array are smaller than it. That is, for each nums[i]
you have to count the number of valid j's
such that j != i
and nums[j] < nums[i]
.
Return the answer in an array.
Example 1:
Input: nums = [8,1,2,2,3] Output: [4,0,1,1,3] Explanation: For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and 3). For nums[1]=1 does not exist any smaller number than it. For nums[2]=2 there exist one smaller number than it (1). For nums[3]=2 there exist one smaller number than it (1). For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).
Example 2:
Input: nums = [6,5,4,8] Output: [2,1,0,3]
Example 3:
Input: nums = [7,7,7,7] Output: [0,0,0,0]
Constraints:
2 <= nums.length <= 500
0 <= nums[i] <= 100
A:
class Solution { public: vector<int> smallerNumbersThanCurrent(vector<int>& nums) { vector<int> copy = nums; sort(copy.begin(), copy.end()); unordered_map<int, int> myMap; for(int i =0;i<copy.size();i++){ int val = copy[i]; if( myMap.find(val) == myMap.end()){ // if not found myMap[val] = i; } } vector<int> res; for(auto k:nums){ res.push_back(myMap[k]); } return res; } };
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