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Algorithm 202: 3 Ways to Sum a Range of Values

NJOKU SAMSON EBERE on March 20, 2020

Given an array of two element as minimum and maximum values ([min, max]), how can you sum up all the values between the min and max values? ran...
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Ganesh Shetty • • Edited

This also works pretty well.

function findSum([min, max]) {
var sum = 0;
while (min <= max) {
sum += min;
min++
}
return sum;
}
console.log(findSum([5, 12]));

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Kay Kleinvogel •

I tried to implement the algorithm myself and came up with a solution that is utilizing the formula for sums by Gauss.

IDK how to explain the equation but it would look like this:

Sum=∑i=1maxi−∑i=1mini=max⋅(max+1)2−min⋅(min+1)2 Sum=\sum_{i=1}^{max}i - \sum_{i=1}^{min}i = \frac{max\cdot (max+1)}{2}-\frac{min\cdot(min+1)}{2}

I implemented this in typescript to see if it works:

const rangeSum = (boundaries: Array<number>): number => {
    const startingPoint = boundaries[0];
    const endPoint = boundaries[1];
    const sum = gaus(endPoint) - gaus(startingPoint - 1);
    return sum;
};

const gaus = (limit: number): number => {
    const gaus_sum = (limit * (limit + 1)) / 2;
    return gaus_sum;
};

The startingPoint - 1 is adjusting that the formula would be excluding the min and we want to include it. In general it seems to be working quiet well.

The advantage here also is that it only has to execute 2 operations no matter how large the span is.
This would decrease the time complexity to O(1). Which is quiet nice.

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NJOKU SAMSON EBERE •

Kay, thank you for your suggestion. This is how I did it with javascript:

function rangeSum(array) {
  let max = array[1];
  let min = array[0];

  let maxManipulation = (max * (max + 1))/2;
  let minManipulation = (min * (min + 1))/2;

  return maxManipulation - minManipulation + array[0];
}

So, I had to add the lower boundary (min value) to the final answer to make it work properly.

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Kay Kleinvogel • • Edited

Yeah. You either have to add the min value or subtract minManipulation(array[0]-1) because otherwise it will exclude the lower boundary. So I corrected the formula to reflect this change (just in case anybody is interested in this).

∑i=minmaxi=∑i=1maxi−∑i=1mini+min∑i=minmaxi=max⋅(max+1)2−min⋅(min+1)2+min \sum_{i=min}^{max}i=\sum_{i=1}^{max}i-\sum_{i=1}^{min}i + min \newline \sum_{i=min}^{max}i= \frac{max\cdot (max+1)}{2}-\frac{min\cdot(min+1)}{2}+min
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NJOKU SAMSON EBERE •

Looking Good 💪