##### Daily Algorithms - 2/13/19
22 months ago
`Happy hacking!`
1
22 months ago
```Saw an interesting whiteboard problem from Improving yesterday.  Check if a given array is balanced where balanced means that there exists a position in which the sum of all numbers on the left is equal to the sum of all the elements on the right.  For example:

[1, 2, 3, 3, 2, 1] is balanced
[1, 2, 3, 4, 5] is not balanced

I don't remember if they dealt with odd arrays such as:
[1, 2, 3, 2, 1] might be balanced?

Assuming the above is unbalanced, here's one O(n) brute-force solution.

let arr1 = [1, 2, 3, 4, 5];
let arr2 = [1, 2, 3, 3, 2, 1];

function checkBalance(arr) {

// Balanced if size is 0 or 1
if (arr.length < 2) {
return true;
}

// Balanced if both values are equal
if (arr.length == 2) {
if (arr == arr) {
return true;
}
}

for (let i = 1; i < arr.length; i ++) {
let leftArr = arr.slice(0, i);
let rightArr = arr.slice(i, arr.length);
let leftSum = leftArr.reduce((a, b) => {return a + b});
let rightSum = rightArr.reduce((a, b) => {return a + b});
if (leftSum == rightSum) {
return true;
}
}
return false;
}

console.log(checkBalance(arr1));
console.log(checkBalance(arr2));```
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