Patterns

Two pointers across two sorted inputs, advancing whichever is smaller

When to useMerging sorted arrays/lists, intersection/union of sorted sequences

iterative

Merge Sorted Arrays

merge_sorted_arrays

Merge Two Sorted Lists

merge_two_sorted_lists

Teleporter Arrays

get_maximum_score

INITIALIZE

i, j for each input; result array

const result = [];
let i = 0, j = 0;

const dummy = new ListNode(0);
let cur = dummy;

const MOD = 10 ** 9 + 7;
let i = 0, j = 0;
let sum1 = 0, sum2 = 0;
let total = 0;

TRAVERSE

while both have elements

while (i < nums1.length && j < nums2.length) {

while (l1 && l2) {

while (i < arr1.length || j < arr2.length) {

[COMPARE]

take smaller into result

  if (nums1[i] <= nums2[j]) result.push(nums1[i++]);
  else result.push(nums2[j++]);

  if (l1.val <= l2.val) { cur.next = l1; l1 = l1.next; }
  else { cur.next = l2; l2 = l2.next; }

  if (arr1[i] === arr2[j]) { total = (total + Math.max(sum1, sum2) + arr1[i]) % MOD; sum1 = 0; sum2 = 0; i++; j++; }
  else if (arr1[i] < arr2[j]) sum1 += arr1[i++];
  else sum2 += arr2[j++];

advance

advance the chosen pointer

—

  cur = cur.next;

—

drain

append leftovers from non-empty side

while (i < nums1.length) result.push(nums1[i++]);
while (j < nums2.length) result.push(nums2[j++]);

cur.next = l1 !== null ? l1 : l2;

// final segment: max(sum1, sum2) captured after loop

RETURN

merged result

return result;

return dummy.next;

return (total + Math.max(sum1, sum2)) % MOD;