<pre class="shiki github-dark shaku"><code>/** * @param {string} str * @returns {string} */function uncompress(str) { // your code here const strStack = []; for (let i = 0; i &lt; str.length; i++) { if(str[i] !== ")") { // push to stack until closing bracket strStack.push(str[i]) } else { let word = ""; let count = ""; let num = ""; // start popping from the stack and keep appending the word while(strStack.length &gt; 0 &amp;&amp; strStack[strStack.length - 1] !== "(") { word = strStack.pop() + word; } strStack.pop(); // pop the "(" // start popping the numerical values which can be in multiple digits while(strStack.length &gt; 0 &amp;&amp; parseInt(strStack[strStack.length - 1])){ num = strStack.pop() + num; } // form the word by repeating itself num times while(+num-- &gt; 0) { count = count + word; } strStack.push(count); } } return strStack.join("");}console.log(uncompress('3(ab2(c))'))</code></pre>

Easy Solution with Explanation

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Given a compressed string, return its original form. For example. uncompress('3(ab)') // 'ababab'uncompress('3(ab2(c))') // 'abccabccabcc' 1. a number…