Related Articles
Add two numbers represented by two arrays
• Difficulty Level : Easy
• Last Updated : 19 Mar, 2021

Given two array A[0….n-1] and B[0….m-1] of size n and m respectively, representing two numbers such that every element of arrays represent a digit. For example, A[] = { 1, 2, 3} and B[] = { 2, 1, 4 } represent 123 and 214 respectively. The task is to find the sum of both the number. In above case, answer is 337.
Examples :

```Input : n = 3, m = 3
a[] = { 1, 2, 3 }
b[] = { 2, 1, 4 }
Output : 337
123 + 214 = 337

Input : n = 4, m = 3
a[] = { 9, 5, 4, 9 }
b[] = { 2, 1, 4 }
Output : 9763```

The idea is to start traversing both the array simultaneously from the end until we reach the 0th index of either of the array. While traversing each elements of array, add element of both the array and carry from the previous sum. Now store the unit digit of the sum and forward carry for the next index sum. While adding 0th index element if the carry left, then append it to beginning of the number.
Below is the illustration of approach: Below is the implementation of this approach:

## C++

 `// CPP program to sum two numbers represented two``// arrays.``#include ``using` `namespace` `std;` `// Return sum of two number represented by the arrays.``// Size of a[] is greater than b[]. It is made sure``// be the wrapper function``int` `calSumUtil(``int` `a[], ``int` `b[], ``int` `n, ``int` `m)``{``    ``// array to store sum.``    ``int` `sum[n];` `    ``int` `i = n - 1, j = m - 1, k = n - 1;` `    ``int` `carry = 0, s = 0;` `    ``// Until we reach beginning of array.``    ``// we are comparing only for second array``    ``// because we have already compare the size``    ``// of array in wrapper function.``    ``while` `(j >= 0) {` `        ``// find sum of corresponding element``        ``// of both arrays.``        ``s = a[i] + b[j] + carry;``        ``sum[k] = (s % 10);` `        ``// Finding carry for next sum.``        ``carry = s / 10;` `        ``k--;``        ``i--;``        ``j--;``    ``}` `    ``// If second array size is less the first``    ``// array size.``    ``while` `(i >= 0) {` `        ``// Add carry to first array elements.``        ``s = a[i] + carry;``        ``sum[k] = (s % 10);``        ``carry = s / 10;` `        ``i--;``        ``k--;``    ``}` `    ``int` `ans = 0;` `    ``// If there is carry on adding 0 index elements.``    ``// append 1 to total sum.``    ``if` `(carry)``        ``ans = 10;` `    ``// Converting array into number.``    ``for` `(``int` `i = 0; i <= n - 1; i++) {``        ``ans += sum[i];``        ``ans *= 10;``    ``}` `    ``return` `ans / 10;``}` `// Wrapper Function``int` `calSum(``int` `a[], ``int` `b[], ``int` `n, ``int` `m)``{``    ``// Making first array which have``    ``// greater number of element``    ``if` `(n >= m)``        ``return` `calSumUtil(a, b, n, m);` `    ``else``        ``return` `calSumUtil(b, a, m, n);``}` `// Driven Program``int` `main()``{``    ``int` `a[] = { 9, 3, 9 };``    ``int` `b[] = { 6, 1 };` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a);``    ``int` `m = ``sizeof``(b) / ``sizeof``(b);` `    ``cout << calSum(a, b, n, m) << endl;` `    ``return` `0;``}`

## Java

 `// Java program to sum two numbers ``// represented two arrays.``import` `java.io.*;` `class` `GFG {` `    ``// Return sum of two number represented by``    ``// the arrays. Size of a[] is greater than``    ``// b[]. It is made sure be the wrapper``    ``// function``    ``static` `int` `calSumUtil(``int` `a[], ``int` `b[],``                                ``int` `n, ``int` `m)``    ``{``        ``// array to store sum.``        ``int``[] sum= ``new` `int``[n];``    ` `        ``int` `i = n - ``1``, j = m - ``1``, k = n - ``1``;``    ` `        ``int` `carry = ``0``, s = ``0``;``    ` `        ``// Until we reach beginning of array.``        ``// we are comparing only for second``        ``// array because we have already compare``        ``// the size of array in wrapper function.``        ``while` `(j >= ``0``)``        ``{``            ``// find sum of corresponding element``            ``// of both array.``            ``s = a[i] + b[j] + carry;``            ` `            ``sum[k] = (s % ``10``);``    ` `            ``// Finding carry for next sum.``            ``carry = s / ``10``;``    ` `            ``k--;``            ``i--;``            ``j--;``        ``}``    ` `        ``// If second array size is less``        ``// the first array size.``        ``while` `(i >= ``0``)``        ``{``            ``// Add carry to first array elements.``            ``s = a[i] + carry;``            ``sum[k] = (s % ``10``);``            ``carry = s / ``10``;``    ` `            ``i--;``            ``k--;``        ``}``    ` `        ``int` `ans = ``0``;``    ` `        ``// If there is carry on adding 0 index``        ``// elements  append 1 to total sum.``        ``if` `(carry == ``1``)``            ``ans = ``10``;``    ` `        ``// Converting array into number.``        ``for` `( i = ``0``; i <= n - ``1``; i++) {``            ``ans += sum[i];``            ``ans *= ``10``;``        ``}``    ` `        ``return` `ans / ``10``;``    ``}``    ` `    ``// Wrapper Function``    ``static` `int` `calSum(``int` `a[], ``int` `b[], ``int` `n,``                                        ``int` `m)``    ``{``        ``// Making first array which have``        ``// greater number of element``        ``if` `(n >= m)``            ``return` `calSumUtil(a, b, n, m);``    ` `        ``else``            ``return` `calSumUtil(b, a, m, n);``    ``}``    ` `    ``/* Driver program to test above function */``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `a[] = { ``9``, ``3``, ``9` `};``            ``int` `b[] = { ``6``, ``1` `};``        ` `            ``int` `n = a.length;``            ``int` `m = b.length;``        ``System.out.println(calSum(a, b, n, m));``    ``}``}` `// This article is contributed by Gitanjali.`

## Python3

 `# Python3 code to sum two numbers``# representer two arrays.` `# Return sum of two number represented``# by the arrays. Size of a[] is greater``# than b[]. It is made sure be the``# wrapper function``def` `calSumUtil( a , b , n , m ):``    ``# array to store sum.``    ``sum` `=` `[``0``] ``*` `n``    ``i ``=` `n ``-` `1``    ``j ``=` `m ``-` `1``    ``k ``=` `n ``-` `1``    ` `    ``carry ``=` `0``    ``s ``=` `0``    ` `    ``# Until we reach beginning of array.``    ``# we are comparing only for second array``    ``# because we have already compare the size``    ``# of array in wrapper function.``    ``while` `j >``=` `0``:` `        ``# find sum of corresponding element``        ``# of both array.``        ``s ``=` `a[i] ``+` `b[j] ``+` `carry``        ``sum``[k] ``=` `(s ``%` `10``)``        ` `        ``# Finding carry for next sum.``        ``carry ``=` `s ``/``/` `10``        ` `        ``k``-``=``1``        ``i``-``=``1``        ``j``-``=``1``    ` `    ``# If second array size is less the first``    ``# array size.``    ``while` `i >``=` `0``:` `        ``# Add carry to first array elements.``        ``s ``=` `a[i] ``+` `carry``        ``sum``[k] ``=` `(s ``%` `10``)``        ``carry ``=` `s ``/``/` `10``        ` `        ``i``-``=``1``        ``k``-``=``1``    ` `    ``ans ``=` `0``    ``# If there is carry on adding 0 index elements.``    ``# append 1 to total sum.``    ``if` `carry:``        ``ans ``=` `10``    ` `    ``# Converting array into number.``    ``for` `i ``in` `range``(n):``        ``ans ``+``=` `sum``[i]``        ``ans ``*``=` `10``    ` `    ``return` `ans ``/``/` `10` `# Wrapper Function``def` `calSum(a, b, n, m ):` `    ``# Making first array which have``    ``# greater number of element``    ``if` `n >``=` `m:``        ``return` `calSumUtil(a, b, n, m)``    ``else``:``        ``return` `calSumUtil(b, a, m, n)` `# Driven Code``a ``=` `[ ``9``, ``3``, ``9` `]``b ``=` `[ ``6``, ``1` `]``n ``=` `len``(a)``m ``=` `len``(b)``print``(calSum(a, b, n, m))` `# This code is contributed by "Sharad_Bhardwaj".`

## C#

 `// C# program to sum two numbers``// represented two arrays.``using` `System;` `class` `GFG {` `    ``// Return sum of two number represented by``    ``// the arrays. Size of a[] is greater than``    ``// b[]. It is made sure be the wrapper``    ``// function``    ``static` `int` `calSumUtil(``int` `[]a, ``int` `[]b,``                                ``int` `n, ``int` `m)``    ``{``        ``// array to store sum.``        ``int``[] sum= ``new` `int``[n];``    ` `        ``int` `i = n - 1, j = m - 1, k = n - 1;``    ` `        ``int` `carry = 0, s = 0;``    ` `        ``// Until we reach beginning of array.``        ``// we are comparing only for second``        ``// array because we have already compare``        ``// the size of array in wrapper function.``        ``while` `(j >= 0)``        ``{``            ``// find sum of corresponding element``            ``// of both array.``            ``s = a[i] + b[j] + carry;``            ` `            ``sum[k] = (s % 10);``    ` `            ``// Finding carry for next sum.``            ``carry = s / 10;``    ` `            ``k--;``            ``i--;``            ``j--;``        ``}``    ` `        ``// If second array size is less``        ``// the first array size.``        ``while` `(i >= 0)``        ``{``            ``// Add carry to first array elements.``            ``s = a[i] + carry;``            ``sum[k] = (s % 10);``            ``carry = s / 10;``    ` `            ``i--;``            ``k--;``        ``}``    ` `        ``int` `ans = 0;``    ` `        ``// If there is carry on adding 0 index``        ``// elements append 1 to total sum.``        ``if` `(carry == 1)``            ``ans = 10;``    ` `        ``// Converting array into number.``        ``for` `( i = 0; i <= n - 1; i++) {``            ``ans += sum[i];``            ``ans *= 10;``        ``}``    ` `        ``return` `ans / 10;``    ``}``    ` `    ``// Wrapper Function``    ``static` `int` `calSum(``int` `[]a, ``int` `[]b, ``int` `n,``                                        ``int` `m)``    ``{``        ``// Making first array which have``        ``// greater number of element``        ``if` `(n >= m)``            ``return` `calSumUtil(a, b, n, m);``    ` `        ``else``            ``return` `calSumUtil(b, a, m, n);``    ``}``    ` `    ``// Driver program``    ``public` `static` `void` `Main()``    ``{``        ``int` `[]a = { 9, 3, 9 };``        ``int` `[]b = { 6, 1 };``        ` `        ``int` `n = a.Length;``    ` `        ``int` `m = b.Length;``        ``Console.WriteLine(calSum(a, b, n, m));``    ``}``}` `// This article is contributed by vt_m.`

## PHP

 `= 0)``    ``{``        ``// find sum of corresponding``        ``// element of both array.``        ``\$s` `= ``\$a``[``\$i``] + ``\$b``[``\$j``] + ``\$carry``;``        ``\$sum``[``\$k``] = (``\$s` `% 10);` `        ``// Finding carry for next sum.``        ``\$carry` `= ``\$s` `/ 10;` `        ``\$k``--;``        ``\$i``--;``        ``\$j``--;``    ``}` `    ``// If second array size is less``    ``// than the first array size.``    ``while` `(``\$i` `>= 0)``    ``{``        ``// Add carry to first array elements.``        ``\$s` `= ``\$a``[``\$i``] + ``\$carry``;``        ``\$sum``[``\$k``] = (``\$s` `% 10);``        ``\$carry` `= ``\$s` `/ 10;` `        ``\$i``--;``        ``\$k``--;``    ``}` `    ``\$ans` `= 0;` `    ``// If there is carry on``    ``// adding 0 index elements.``    ``// append 1 to total sum.``    ``if` `(``\$carry``)``        ``\$ans` `= 10;` `    ``// Converting array into number.``    ``for` `( ``\$i` `= 0; ``\$i` `<= ``\$n` `- 1; ``\$i``++)``    ``{``        ``\$ans` `+= ``\$sum``[``\$i``];``        ``\$ans` `*= 10;``    ``}` `    ``return` `\$ans` `/ 10;``}` `// Wrapper Function``function` `calSum( ``\$a``, ``\$b``, ``\$n``, ``\$m``)``{``    ``// Making first array which have``    ``// greater number of element``    ``if` `(``\$n` `>= ``\$m``)``        ``return` `calSumUtil(``\$a``, ``\$b``, ``\$n``, ``\$m``);` `    ``else``        ``return` `calSumUtil(``\$b``, ``\$a``, ``\$m``, ``\$n``);``}` `// Driven Code``\$a` `= ``array``( 9, 3, 9 );``\$b` `= ``array``( 6, 1 );` `\$n` `= ``count``(``\$a``);``\$m` `= ``count``(``\$b``);` `echo` `calSum(``\$a``, ``\$b``, ``\$n``, ``\$m``);` `// This article is contributed by anuj_67.``?>`

## Javascript

 ``

Output :

`1000`

Time Complexity: O(n + m)

Auxiliary Space: O(max(n, m))

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up