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Given an array which is a collection of non-negative number represented as an array of digits, add 1 to the number (increment the number represented by the digits ). The digits are stored such that the most significant digit is the first element of the array.

To add 1 to the number represented by digits

Given array from the end, addition means rounding of the last no 4 to 5.

If the last elements 9, make it 0 and carry = 1.

For the next iteration check carry and if it adds to 10, do the same as step 2.

After adding carry, make carry = 0 for the next iteration.

If the vectors add and increase the vector size, append 1 in the beginning.

Suppose an array contains elements [7, 6, 3, 4] then the array represents decimal number 1234 and hence adding 1 to this would result in 7635. So new array will be [7, 6, 3, 5].

Input: [7, 6, 9, 9] Output: [7, 7, 0, 0] Input: [4, 1, 7, 8, 9] Output: [4, 1, 7, 9, 0]

**Explanation** Add 1 to the last element of the array, if it is less than 9. If the element is 9, then make it 0 and recurse for the remaining element of the array.

#include <iostream> using namespace std; void sum(int arr[], int n) { int i = n; if(arr[i] < 9) { arr[i] = arr[i] + 1; return; } arr[i] = 0; i--; sum(arr, i); if(arr[0] > 0) { cout << arr[0] << ", "; } for(int i = 1; i <= n; i++) { cout << arr[i]; if(i < n) { cout << ", "; } } } int main() { int n = 4; int arr[] = {4, 1, 7, 8, 9}; sum(arr, n); return 0; }

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