minimum-value-to-equal-array.md

Requirements

Given a list of integers, return the number of times it takes produce a list of equal integers by incrementing n - 1 elements by one.

Strategy

Maths!

Justification of strategy

The largest number in the list, x, is greater than or equal to the product of s, the smallest number, 1, the number we increment the smallest number by, and m, the number of times we increment, and the value we're after.  Refactoring, we find that the product of m and the number of elements we increment equals the difference between the sum of all elements and the product of the largest element and the number of elements.  = sum-x*n >= sum-(minNum+m)n) Refactoring again, we get *n) Just solve for

Define test cases

• minToEqual([1, 2, 3]) should return 3
• minToEqual([1, 50000]) should return 49999

Implementation skeleton

• a: solve for sum of elements
• b: solve for minimum value
• c: solve for length of list
• return a - (b * c)

Actual implementation

function minMoves(list) {
  var sum = list.reduce((a, b) => a + b)
  var minNum = Math.min(...list)
  var n = list.length;
  return sum-(minNum)*n
}

Execute test cases

Big-O Analysis

O(n), Linear because of reduce method.

Optimization (if applicable)