[Algorithm] Find Nth smallest value from Array

Q1: Find the smallest value from array: 

function findMin(arr) {
  let min = arr[0];

  for (let i = 1; i < arr.length; i++) {
    if (arr[i] < min) {
      min = arr[i];
    }
  }

  return min; // 1
}

O(n), cannot be improved anymore, because we have to loop though the array once.

 

Q2: Find 2nd smallest value, naive solution:

function findSndMin(arr) {
  let min = arr[0];
  let min2 = arr[1];

  for (let i = 0; i < arr.length; i++) {
    if (arr[i] < min) {
      min2 = min;
      min = arr[i];
    } else if (arr[i] !== min && arr[i] < min2) {
      min2 = arr[i];
    }
  }

  return min2; // 2
}

 

Q3: Find nth smallest value:

1. Sorting cannot be the best solution, because it do more works, we only care Nth, means I don't care 0...n-1 is sorted or not or n +1....arr.length is sorted or not.

2. Patition: avg can achieve n + n/2 + n/4 + n/8 .... = 2n ~~ O(n), worse: O(n^2)

function findNthMin(arr, m) {
  const pivot = arr[0];
  let smaller = [];
  let larger = [];

  for (let i = 1; i < arr.length; i++) {
    if (arr[i] < pivot) {
      smaller.push(arr[i]);
    } else {
      larger.push(arr[i]);
    }
  }

  smaller.push(pivot);

  arr = [...smaller, ...larger];

  if (m > smaller.length) {
    return findNthMin(larger, m - smaller.length);
  } else if (m < smaller.length) {
    return findNthMin(smaller, m);
  } else {
    return arr[m - 1];
  }
}

const data = [3, 1, 5, 7, 2, 8];
const res = findNthMin(data, 4);
console.log(res);

 

Partition vs Heap:

Why here Heap is not a good solution, because we only need Nth samllest value, we don't care  0..n-1, whether those need to be sorted or not. Heap doing more stuff then necessary.

 

But if the question change to "Find first K smallest value", then Heap is a good solution.

Mean while we need to be carefull that since we just need K samllest, not all N, then we don't need to add everything into the Heap.

if (h,size() >= K) {
  if (h.peek() > val(i)) {
    h.pop()
    h.add(val(i))
  }  
} else {
    h.add(val(i))
}