Find missing number between 1 to 100 - c

Question

This question has been asked here on SO before with below code
find3missing(int* array)
{
int newarray[100] = {0};
For i = 0 to 99
++newarray[array[i]] ;
For i = 0 to 99
If newarray[i] != 1
Cout << “the missing number is ” << i+1 << endl ;
}
But when I checked this code, it doesn't seem to work. Suppose I have an array of {1,2,6}. The output should be 3,4,5 but with the code above I get 1,4,5,6 instead. Below is my implementation of pseudo code with array size 6.
main()
{
int a[6]={1,2,6};
int tmp[6]={0},i;
for(i=0;i<6;i++)
{
++tmp[a[i]];
}
for(i=0;i<6;i++)
{
if(tmp[i]!=1)
{
printf("%d",i+1);
}
}
}
Is this the right code?

Answer 1

This ++newarray[array[i]] should be ++newarray[array[i] - 1]. This because you are interested in a sequence of 1-100 numbers, so no 0, but C arrays are 0 based. If you then look at the cout: the missing number is ” << i+1 here you "unshift" the number by adding 1.
There is another problem: you should pass the number of elements of the array, something like:
find3missing(int* array, int length) {
int newarray[100] = {0};
for (int i = 0; i < length; i++) {
++newarray[array[i] - 1] ;
}

Answer 2

C/C++ arrays are zero based, as A[i] is equivalent to *(A+i). So change ++newarray[array[i]] to ++newarray[array[i]-1]. Also use malloc, free and memset to use an array of dynamic size.
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
void find3missing(int* pArray, size_t size, int min, int max){
int* newarray;
int i;
unsigned int j;
int range = max - min;
if(range < 0)
return;
newarray = (int*) malloc(range*sizeof(int)); // allocate enough memory
memset(newarray,0,range*sizeof(int)); // set that block to zero
for(j = 0; j < size; ++j){
++newarray[pArray[j]-min];
}
for(i = 0; i < range; ++i){
if(!newarray[i])
printf("%d is missing!\n",min+i);
}
free(newarray);
}
int main(){
int test[] = {1,3,6};
find3missing(test,sizeof(test)/sizeof(int),1,6);
return 0;
}
Please note that this solution is very inefficient if your array is sorted. In this case have a look at Jimmy Gustafsson's answer.

Answer 3

This algoritm will be quite simple, since you're using a sorted array. Simply check if the current value +1 equals the nextvalue like below:
find3missing(){
int array[arraySize]; // the array with integers
for(i=0;i<arraySize;i++)
if(array[i]+1 != array[i+1]) // if value array[i]+1 is not equal the next index
// value, then it's a missing number
printf("A missing number: %i", i+1);
}