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howManyNumbersAreSmallerThanTheCurrentNumber.js
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howManyNumbersAreSmallerThanTheCurrentNumber.js
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// LeetCode #1365. How Many Numbers Are Smaller Than The Current Number
/* Instructions
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
*/
// Solution
const smallerNumbersThanCurrent = nums => {
// instantiate empty array to hold our output
let output = []
// iterate through the nums array
for (let i = 0; i < nums.length; i++) {
let count = 0;
let j = 0
// loop through every next element in array
while (j < nums.length) {
if (nums[i] > nums[j]) {
count++
j++
} else {
j++
}
}
output.push(count)
}
return output
}