/**
 * 
 * Ashish Patel
 * e: ashishsushilPatel@gmail.com
 * w: https://ashish.me
 *
 */

/* 
 * 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].
 * 
 * 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).
 */

function howManyNumbersAreSmallerThanTheCurrentNumber(value) {
  let result = new Array(value.length).fill(0)
  for (let i = 0; i < value.length; i++) {
    for (let j = 0; j < value.length; j++) {
      if(value[i] !== value[j] && value[j] < value[i]){
        result[i] += 1
      }
    }    
  }
  return result
}

test('how Many Numbers Are Smaller Than The Current Number', () => {
  expect(howManyNumbersAreSmallerThanTheCurrentNumber([8,1,2,2,3])).toEqual([4,0,1,1,3])
});