LeetCode 1365. How Many Numbers Are Smaller Than the Current Number

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Question

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

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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).

Input: nums = [6,5,4,8]
Output: [2,1,0,3]

Solution

It’s a sorting question. We have to know the position of each element in array. So, first method I used would be sort it and used HashMap to track the relationship. In addition, we can use bucket sort to reduce the time complexity.

Code

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// time:O(nlogn) space:O(n)
    public int[] smallerNumbersThanCurrent(int[] nums) {
        if (nums == null || nums.length == 0) return new int[]{};
        int n = nums.length;
        int[] res = new int[n];
        int[] copy = nums.clone();
        HashMap<Integer, Integer> map = new HashMap<>();
        Arrays.sort(copy);
        for (int i = 0; i < n; i++) {
            if (!map.containsKey(copy[i])) 
                map.put(copy[i], i);
        }
        for (int i = 0; i < n; i++) {
            res[i] = map.get(nums[i]);
        }
        return res;
    }

    // bucket sort + prefix Sum
    // time:O(n) space:O(1)
    public int[] smallerNumbersThanCurrent2(int[] nums) {
        if (nums == null || nums.length == 0) return new int[]{};
        int n = nums.length;
        int[] res = new int[n];
        int size = 100;
        int[] count = new int[size + 1];
        for (int num : nums) count[num]++;
        for (int i = 1; i <= size; i++) {
            count[i] += count[i - 1];
        }
        for (int i = 0; i < n; i++) {

            if (nums[i] - 1 >= 0)  
                res[i] = count[nums[i] - 1];
        }
        return res;
    }