<a href="https://leetcode.com/problems/count-and-say/editorial/">Leetcode 38. Count and Say</a>
Overview
Given a mutable variable <code>read</code> as the based case <code>'1'</code> it will be updated to store the read of the previous case in the while loop based on n.
<ul>
<li>in the while loop:
<ul>
<li>(1) instantiate 3 mutable variables:
(a) <code>count</code>: to store the count of the say (of the digit), (b) <code>say</code>: the digit (c) <code>readSegs</code>: the accumulation of the read string for current while loop iteration</li>
<li>(2) iterate through the current <code>read</code> string, each time to compare whether the <code>currentSay</code> as same as the <code>say</code>. If it is, add one to count. if it is not, update the <code>readSeg</code>, replace the <code>say</code> and reset the <code>count</code> to 1.</li>
</ul>
</li>
</ul>
Complexity Analysis
<ul>
<li>time complexity: O(4^{n/3})</li>
<li>space complexity: O(4^{n/3})</li>
</ul>
<pre class="shiki github-dark shaku"><code>/** * @param {number} n - integer * @returns {string} */function getNthNum(n) { // your code here let read = '1'; while ( n &gt; 1 ) { let count = 0; let say = read[0]; let readSegs = ''; for (let i = 0; i &lt; read.length; i++ ) { let currSay = read[i]; if ( currSay === say) { count++; } else { readSegs = `${readSegs}${count}${say}`; count = 1; say = currSay; } } read = `${readSegs}${count}${say}`; n--; } return read;}console.log(getNthNum(1))// '1'console.log(getNthNum(2))// '11'console.log(getNthNum(3))// '21'console.log(getNthNum(4))// '1211'</code></pre>