Write your own strcmp that ignores cases


Write a modified strcmp function which ignores cases and returns -1 if s1 < s2, 0 if s1 = s2, else returns 1. For example, your strcmp should consider "GeeksforGeeks" and "geeksforgeeks" as same string.

Source: Microsoft Interview Set 5

Following solution assumes that characters are represented using ASCII representation, i.e., codes for ‘a’, ‘b’, ‘c’, … ‘z’ are 97, 98, 99, … 122 respectively. And codes for ‘A’, “B”, ‘C’, … ‘Z’ are 65, 66, … 95 respectively.

Following are the detailed steps.
1) Iterate through every character of both strings and do following for each character.
a) If str1[i] is same as str2[i], then continue.
b) If inverting the 6th least significant bit of str1[i] makes it same as str2[i], then continue. For example, if str1[i] is 65, then inverting the 6th bit will make it 97. And if str1[i] is 97, then inverting the 6th bit will make it 65.
c) If any of the above two conditions is not true, then break.
2) Compare the last (or first mismatching in case of not same) characters.

#include <stdio.h>

/* implementation of strcmp that ignores cases */
int ic_strcmp(char *s1, char *s2)
    int i;
    for (i = 0; s1[i] && s2[i]; ++i)
        /* If characters are same or inverting the 
           6th bit makes them same */
        if (s1[i] == s2[i] || (s1[i] ^ 32) == s2[i])

    /* Compare the last (or first mismatching in 
       case of not same) characters */
    if (s1[i] == s2[i])
        return 0;

    // Set the 6th bit in both, then compare
    if ((s1[i] | 32) < (s2[i] | 32)) 
        return -1;
    return 1;

// Driver program to test above function
int main(void)
    printf("ret: %d\n", ic_strcmp("Geeks", "apple"));
    printf("ret: %d\n", ic_strcmp("", "ABCD"));
    printf("ret: %d\n", ic_strcmp("ABCD", "z"));
    printf("ret: %d\n", ic_strcmp("ABCD", "abcdEghe"));
    printf("ret: %d\n", ic_strcmp("GeeksForGeeks", "gEEksFORGeEKs"));
    printf("ret: %d\n", ic_strcmp("GeeksForGeeks", "geeksForGeeks"));
    return 0;


ret: 1
ret: -1
ret: -1
ret: -1
ret: 0
ret: 0

This article is compiled by Narendra Kangralkar. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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