The Brackets challenge from Codility was a fun exercise.

A solution with O(N) complexity requires a single run of the array.

At each step, we operate on a stack following three cases:

- If the stack is empty and we the current element is a closing bracket of any kind, then the string is not properly nested
- If the stack is not empty and we are placing a closing bracket, then if the closing bracket matches the last opened bracket in the stack, then we eliminate the last opening bracket from the stack
- Otherwise, we add to the stack the element if it is an opening bracket. If we are trying to push a closing bracket that did not match, again we have the conclusion that the string is not properly nested.

## Solution in Javascript

```
function solution(S) {
// write your code in JavaScript (Node.js 8.9.4)
let progress = true
let index = 0
let stack = []
let error = false
while (index < S.length) {
if (stack.length === 0) {
if (S[index] === ')' || S[index] === ']' || S[index] === '}') {
return 0
} else {
stack.push(S[index])
}
}
else if (is_properly_nested(stack[stack.length-1],S[index])) {
stack.pop(stack[stack.length-1])
} else {
if (S[index] === ')' || S[index] === ']' || S[index] === '}') {
return 0
}
else {
stack.push(S[index])
}
}
index ++
}
if (stack.length === 0) {
return 1
} else {
return 0
}
}
function is_properly_nested(a,b){
if ( (a === '(' && b === ')') || (a === '[' && b === ']') || (a === '{' && b === '}')) {
return true
}
else {
return false
}
```

Here is the report. Let me know if you can think of a better solution!

Here is my solution

function solution(S) {

var S = S.split(”)

var stack = []

var arr = {

“)” : “(“,

“}” : “{“,

“]” : “[”

}

var leftBracket = [“(“, “{“, “[“]

for(let i = 0; i

~~0 ? 0 : 1~~}