11. Add Two Numbers
mediumAsked at SwiggyAdd two numbers stored as digits in linked lists in reverse order.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
Two non-empty linked lists each represent a non-negative integer; digits are stored in reverse order, one digit per node. Add the two numbers and return the sum as a new linked list in the same format.
Constraints
1 <= nodes per list <= 1000 <= node.val <= 9No leading zeros except the number 0 itself
Examples
Example 1
l1=[2,4,3], l2=[5,6,4][7,0,8] (342+465=807)Example 2
l1=[9,9,9], l2=[1][0,0,0,1]Approaches
1. Convert to BigInt
Reassemble both numbers, add, split back into digits.
- Time
- O(n)
- Space
- O(n)
function num(l){let s='';for(;l;l=l.next)s=l.val+s;return BigInt(s);}
let sum=(num(l1)+num(l2)).toString();
let head=null;
for (const c of sum) head={val:+c,next:head};
return head;Tradeoff:
2. Single pass with carry
Walk both lists in lockstep, summing digits plus carry into new nodes. Continue while either list has nodes or the carry is non-zero.
- Time
- O(max(m,n))
- Space
- O(max(m,n))
function addTwoNumbers(l1, l2) {
const dummy = { val: 0, next: null };
let curr = dummy, carry = 0;
while (l1 || l2 || carry) {
const a = l1 ? l1.val : 0;
const b = l2 ? l2.val : 0;
const sum = a + b + carry;
carry = Math.floor(sum / 10);
curr.next = { val: sum % 10, next: null };
curr = curr.next;
if (l1) l1 = l1.next;
if (l2) l2 = l2.next;
}
return dummy.next;
}Tradeoff:
Swiggy-specific tips
Swiggy weighs your handling of the trailing carry edge case heavily; missing the while-carry loop is the fastest way to fail their real-time payment-rounding analogue.
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