function mergeSorted(a, b) {
  const out = [];
  let i = 0;
  let j = 0;
  while (i < a.length && j < b.length) {
    out.push(a[i] <= b[j] ? a[i++] : b[j++]);
  }
  while (i < a.length) out.push(a[i++]);
  while (j < b.length) out.push(b[j++]);
  return out;
}
mergeSorted([1, 3, 5], [2, 4, 6]); // [1, 2, 3, 4, 5, 6]

Two pointers give O(n + m), while a.concat(b).sort() is O(n log n). Use <= to stay stable and handle duplicates correctly.

sandbox.js

function mergeSorted(a, b) {
  // TODO
}

Output

Hit “Run tests” to check your solution.

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JavaScriptHard15/15

Write mergeSorted(a, b) merging two sorted arrays into one sorted array, without sort()

By FrontendPro Editorial Team Updated 8/7/2026

Explore the JavaScript interview guide

Examples

Write mergeSorted so that the following cases hold.

InputmergeSorted([1,3,5], [2,4,6])
Output[1,2,3,4,5,6]
InputmergeSorted([], [1,2])
Output[1,2]
InputmergeSorted([1,1], [1])
Output[1,1,1]

1 more hidden test cases - 4 in total will be graded.

Answer

function mergeSorted(a, b) {
  const out = [];
  let i = 0;
  let j = 0;
  while (i < a.length && j < b.length) {
    out.push(a[i] <= b[j] ? a[i++] : b[j++]);
  }
  while (i < a.length) out.push(a[i++]);
  while (j < b.length) out.push(b[j++]);
  return out;
}
mergeSorted([1, 3, 5], [2, 4, 6]); // [1, 2, 3, 4, 5, 6]

Two pointers give O(n + m), while a.concat(b).sort() is O(n log n). Use <= to stay stable and handle duplicates correctly.

sandbox.js

function mergeSorted(a, b) {
  // TODO
}

Output

Hit “Run tests” to check your solution.

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