Solution based on left bisection
<pre class="shiki github-dark shaku"><code>function elementBefore(arr, target){ let left = 0, right = arr.length; while(right &gt; left) { let mid = left + Math.floor((right - left) / 2); if (arr[mid] &gt;= target) { right = mid; } else { left = mid + 1; } } if ((left &lt; arr.length &amp;&amp; left &gt;= 1) &amp;&amp; arr[left] === target) { return arr[left - 1]; } else { return undefined; }}</code></pre>

LEFT BISECTION BASED SOLUTION

This is a variation of 37. implement Binary Search (unique). Your are given a sorted ascending array of number, but might have duplicates, you are ask…