CareerCup

use dynamic array(malloc and free)
if the array size does not increase then
1. do inplace addition of the values from array[last_element]
if the array size increases then
1.copy the array values in temp variable {9,9,9}---> temp = 999.
2.free the array
3.dynamically allocate the array with size length+1
4.add 1 to 999 and copy it to the new array as {1,0,0,0}

If we are interested in the final result, then the we can achieve the result without conversion. The code is given below

#include <stdio.h>
#include <stdlib.h>

int main()
{
	int a[] = {7,3,5,3,9};
	int i;
	
	// Instead of adding 1 separately, set carry initially 
	int carry = 1; 
	
	int size = sizeof(a)/sizeof(a[0]);
	int* result = (int*)malloc((size+1)*sizeof(int));
	for(i=size-1;i>=0;i--)
	{
		result[i+1] = (a[i]+carry)%10;
		carry = (a[i]+carry)/10;
	}
	result[0] = carry;
	for(i=0;i<size+1;i++)
		printf("%d",result[i]);
	printf("\n");
	return 0;
}

Can i get a php program for the above question

am i missing anything


<?php
$ss = array(2,3,4);
$count = count($ss);
$sum = 0;
foreach($ss as $key => $item){
$sum += $item * (10 to the power ($count-$key));
}

$final = $sum +1;

$count_integer = lenght($final);

for ($i = 0; $i < $count_integer; $i++){
$new_array[] = $final % 10;
$final = ceil($final/10);
}
?>

import java.util.Arrays;

public class ArrayToNumber {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		//int [] inputArray = {9,9,9};
		int [] inputArray = {7,3,5,3,9};
		System.out.println( "out put:"+ Arrays.toString(inputNumber(inputArray)));

	}
	/**
	 * 
		For example input = {7,3,5,3,9} convert this to number 73539, 
		add 1 so it becomes 73540 and convert to array {7,3,5,4,0}. 
		Array can be of any length, so you can't always represent array in form 
		of in-built number format. So you have to do this summation in-place. 
		Also, how would you increase array size in-case input = {9,9,9} 
		so output = (1,0,0,0} 
		Assume, all elements of arrays are between 0 and 9.
	 */
	public static int [] inputNumber(int [] inputArray){
		int integerValue=0;
		int [] outputArray = new int[inputArray.length];
		int [] overFlowOutputArray = null;		
		//Converting array of elements into integer number
		for(int i=0; i<inputArray.length;i++){
			integerValue=integerValue*10+inputArray[i];
		}
		//Adding one to the integer number which got from previous step
		integerValue = integerValue+1;		
		//Converting integer number back array
		for(int j=inputArray.length-1; integerValue>0;j--){ 
			//Check the array need to be increased and return the array
			if(j==-1){
				overFlowOutputArray = new int[inputArray.length+1];
				System.arraycopy(outputArray, 0, overFlowOutputArray, 1, inputArray.length);
				overFlowOutputArray[0] = integerValue%10;
				return overFlowOutputArray;
			}
			outputArray[j] = integerValue%10;
			integerValue = integerValue/10;	
		}
		return outputArray;
	}
}

php code for the above problem

<?php
$input = array(9,9,9);
$number ='';
foreach($input as $val ) {
 $number .=$val;
}
$result = str_split($number+1);

print_r ($result);
?>

import java.util.Stack;

public class ArrayElementToNumber {

	public static void main(String[] args) {
		System.out.println("Enter the length of array");
		int[] arr = { 7, 5, 3, 2, 9,5,2,3
				};
		Stack<Integer> stack = new Stack<Integer>();

		int i = 0;
		int n = 1;
		while (i <= arr.length - 1) {
			stack.push(arr[arr.length - 1 - i] * n);
			n = n * 10;
			i++;
		}
		int x = 0;
		int result = 0;
		while (x < stack.capacity()) {
			int number = stack.pop();
			result = number + result;
			System.out.println(result);
		}
	
	}
}

#include<stdio.h>
#include<conio.h>
int main(){
int n,a[20],b[20],i,count=0;
long num=0,num1;
clrscr();
printf("Enter number of elements in the array: ");
scanf(" %d",&n);
for(i=0;i<n;i++){
printf("Enter the %d element of array: ",i+1);
scanf(" %d",&a[i]);
num=num*10+a[i];
}
printf("\nThe elements inside the array is: ");
for(i=0;i<n;i++)
printf("%d\t",a[i]);
num=num+1;
num1=num;
while(num/10!=0){
count=count+1;
num=num/10;
}count++;
n=count;
for(i=count-1;i>=0;i--){
b[i]=num1%10;
num1=num1/10;
}
printf("\nThe new array is: ");
for(i=0;i<n;i++)
printf(" %d",b[i]);


getch();
return 0;
}

Language: JavaScript
Assumptions: arr is an integer array, num is an integer

function addOneToArray( arr, num ) {
	var tmp = arr.slice(0);
	( (+(tmp.join("")) + num ) + "").split("").forEach( function(el, i) { tmp[i] = +el; } );
	return tmp;
}

There is no need to convert the array to numeric
simply scan the array from the end , until the first letter that is not '9' ( say *cp)is reached
increment *cp, then set elements after cp to '0'
if (cp<array[0]) free and allocate a new array

#include<stdio.h>
int display(int a[],int n)
{
int i;
for(i=0;i<n;i++)
printf("\n%d",a[i]);
return 0;
}
int shift(int a[],int n)
{
int i;
for(i=n;i>0;i--)
a[i]=a[i-1];
a[0]=1;
return n+1;
}
int main()
{
int i,a[10],n;
printf("enter the number of elements to be inserted in the array:\n");
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(i=n-1;i>=-1;i--)
{
if(i<0){
n=shift(a,n);break;}
if(((a[i]+1)/10)<1){
a[i]=a[i]+1;
break;}
else
a[i]=(a[i]+1)%10;
}
display(a,n);
return 0;
}

Simple perl program for this

#! /usr/local/bin/perl -w
use strict;

my @input_array = (8,7,9);
my $number;

foreach my $value (@input_array){
$number = $number.$value;
}

$number = $number + 1;
my @final_array = split(//,$number);

foreach my $val (@final_array){
print "$val\n";
}