atoi function in C

Question

Question in C : given a number in character array , find the biggest two digit number (group of two digits should be checked from left to right) with the condition that the number cannot have same two digits . eg. input: 64998434 , output: 84 (not 99).

Solution : atoi expects string not character array . therefore just increase the array length by 1 and initialize it with terminating character as written below

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

int main()
{
    char no[] = "64998434";
    int big = 0, temp, j = 0, length;

    length = strlen(no);
    //char str[2] = "";  // not working
    char str[3] = "\0";  // working

    if (length <= 0 || length == 1)
    {
        printf("Invalid input");
        return 0;
    }
    else if (length == 2)
    {
        printf("%d", atoi(no));
        return 0;
    }


    //loop to get two digits from input array and store in another array to convert into 
    //integer and than save and check for bigger number

    for (int i = 0; i < length; i += 2)
    {
        str[j] = no[i];
        str[j + 1] = no[i + 1];
        if (str[0] != str[1])
        {
            temp = atoi(str);
            if (temp > big)
            {
                big = temp;
            }
        }
        j = 0;
    }

    if (big != 0)
    {
        printf("%d\n", big);
        return 0;
    }
}

Answer 1

str<\/code> is an array of length 2<\/code> . You store potentially non-null characters in both of the array slots ( j<\/code> is always 0<\/code> in your code):

    str[j] = no[i];
    str[j + 1] = no[i + 1];

Answer 2

Edit:<\/strong> to clarify , number was supposed to be searched from left to right in a group of two digits , therefore output should be 84 not 98.(since my question was not properly understandable)

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

int convert(char *s, int l)
{
    // Initialize a variable
    int num = 0;
    // int n = l;

    // Iterate till length of the string
    for (int i = 0; i < l; i++)

        // Subtract 48 from the current digit
        num = num * 10 + (s[i] - 48);

    return num;
}

int main()
{
    char no[] = "64998434";
    int big = 0, temp, j = 0, length, num, num2;
    for (int i = 0; no[i] != '\0'; i++)
    {
        length = i;
    }

    char str[2] = "";

    if (length <= 0 || length == 1)
    {
        printf("Invalid input");
        return 0;
    }
    else if (length == 2)
    {
        // printf("%d", atoi(no));
        // return 0;
        num2 = convert(no, length);
        printf("%d", num2);
    }

    for (int i = 0; i < length; i += 2)
    {
        str[j] = no[i];
        str[j + 1] = no[i + 1];
        if (str[0] != str[1])
        {
            temp = convert(str, 2);
            if (temp > big)
            {
                big = temp;
            }
        }
        j = 0;
    }

    if (big != 0)
    {
        printf("%d", big);
        return 0;
    }
}

Answer 3

atoi<\/code> does not seem required for this task.

#include <stdio.h>
#include <string.h>

int main() {
    char str[] = "649984343497864236666784328253489";
    int i, imax = -1, len = strlen(str);
    for (i = 0; i + 2 <= len; i += 2) {
        if (str[i] != str[i + 1] && (imax < 0 || strncmp(str + i, str + imax, 2) > 0)) {
            imax = i;
        }
    }
    if (imax < 0)
        printf("no solution\n");
    else
        printf("%.2s\n", str + imax);
    return 0;
}

Answer 4

Seems like a single pass should do it.

// Return -1 on error
// Otherwise return the value of the biggest pair
int biggest_pair(const char *s) {
  int biggest = -1;
  while (s[0]) {
    // This and next character a digit?
    if (!(s[0] >= '0' && s[0] <= '9' && s[1] >= '0' && s[1] <= '9')) {
      return 1;
    }
    int test = (s[0] - '0')*10 + s[1] - '0';
    if (s[0] != s[1] && test > biggest) {
      biggest = test;
    }
    s += 2;
  }
  return test;
}