How to write the code with less time complexity for finding the missing element in given array range?

Question

My function should return the missing element in a given array range. So i first sorted the array and checked if the difference between i and i+1 is not equal to 1, i\'m returni

Answer 1

Well, from the question (as it's supposed to return a single number) and all the existing solution (examples at least), it looks like list is unique. For that case I think we can sumthe entire array and then subtracting with the expected sum between those numbers will generate the output.

sum of the N natural numbers

1 + 2 + ....... + i + ... + n we can evaluate by n * (n+1) / 2

now assume, in our array min is i and max is n

so to evaluate i + (i+1) + ..... + n we can

A = 1 + 2 + ..... + (i-1) + i + (i+1) + .... n (i.e. n*(n+1)/2)

B = 1 + 2 + ..... + (i-1) and

C = A - B will give us the sum of (i + (i+1) + ... + n)

Now, we can iterate the array once and evaluate the actual sum (assume D), and C - D will give us the missing number.

Let's create the same with each step at first (not optimal for performance, but more readable) then we will try to do in a single iteration

let input1 = [2, 3, 1, 5],
    input2 = [2, 3, 1, 5, 4, 6, 7, 9, 10],
    input3 = [3, 4, 5, 6, 8];

let sumNatural = n => n * (n + 1) / 2;

function findMissing(array) {
  let min = Math.min(...array),
      max = Math.max(...array),
      sum = array.reduce((a,b) => a+b),
      expectedSum = sumNatural(max) - sumNatural(min - 1);
      return expectedSum - sum;
}

console.log('Missing in Input1: ', findMissing(input1));
console.log('Missing in Input2: ', findMissing(input2));
console.log('Missing in Input3: ', findMissing(input3));

Now, lets try doing all in a single iteration (as we were iterating 3 times for max, min and sum)

let input1 = [2, 3, 1, 5],
    input2 = [2, 3, 1, 5, 4, 6, 7, 9, 10],
    input3 = [3, 4, 5, 6, 8];

let sumNatural = n => n * (n + 1) / 2;

function findMissing(array) {
  let min = array[0],
      max = min,
      sum = min,
      expectedSum;
  // assuming the array length will be minimum 2
  // in order to have a missing number
  for(let idx = 1;idx < array.length; idx++) {
    let each = array[idx];
    min = Math.min(each, min); // or each < min ? each : min;
    max = Math.max(each, max); // or each > max ? each : max;
    sum+=each; 
  }

  expectedSum = sumNatural(max) - sumNatural(min - 1);
  return expectedSum - sum;
}

console.log('Missing in Input1: ', findMissing(input1));
console.log('Missing in Input2: ', findMissing(input2));
console.log('Missing in Input3: ', findMissing(input3));