Introduction
Here we are given an array, now we have to traverse the array and check for every index element , how many numbers are smaller than it. And we have to count it and store it in a new array. Like nums = [8,1,2,2,3] , Output: [4,0,1,1,3].
Examples
Steps
- take new array and counter variable
- run a nested for loop.
- Both starts from 0 and ends in nums.length
- if( i == j), then skip
- else if( nums[i] > nums[j]) , count++;
- arr[i] = count;
- return arr;
Hints
JavaCode
class Solution {
public int[] smallerNumbersThanCurrent(int[] nums) {
int arr[] = new int[nums.length];
for(int i = 0; i<nums.length; i++){
int count = 0;
for(int j=0; j< nums.length; j++){
if ( i == j){
continue;
}
else if(nums[i] > nums[j]){
count++;
}
}
arr[i] = count;
}
return arr;
}
}
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