<pre class="shiki github-dark shaku"><code>/** * @param {string} str * @return {string} */function compress(str) { if(str.length == 1) return str; let l = 0; let r = 1; let res = ''; let counter = 1; while(l &lt; str.length) { if(str[l] === str[r]) { counter++; r++; continue; } const strTemp = counter === 1 ? str[l] : str[l] + counter; res += strTemp; counter = 1; l = r; r++; } return res;}</code></pre>

Multiple Pointers solution

Given a string, compress the repeating letters with count number compress('a') // 'a'compress('aa') // 'a2'compress('aaa') // 'a3'compress('aaab') // …