Cisco Systems Interview Question for Software Engineers


Country: India
Interview Type: Written Test




Comment hidden because of low score. Click to expand.
2
of 2 vote

for( int i = a.length-1, j = b.length-1; i >= 0 && j >= 0 && a[i] == b[j]; i--;j--);

if ( i < a.length-1 ) return a.substring(++i);
return "";

- Sunny September 15, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

- O(n^2)
 - Given s1 and s2
 - For each substring(i, s2.length) check if substring is in s1
  - if so print substring

- O(n) #on average
 - Given s1 and s2
 - Create suffix hash of s1
  - Ex: s1 = 'abc'
   - suffix hash = { 'a': 'a', 'abc' : 'abc': 'b': 'b', 'bc': 'bc': 'c': 'c' }
  - For each suffix in s2, check hash if exist print

- tnutty2k8 September 15, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

s1.reverse();
s2.reverse();
for (int i = 0; i < s1.length()&&s2.length(); i++) {
    if (s1[i] == s2[i]) {
        System.out.println(s1[i]);
    }
    else {
        break;
    }
}

- harrypotter0 September 16, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

An update to the above code ::

s1.reverse();
s2.reverse();
string s;
for (int i = 0; i < s1.length()&&s2.length(); i++) {
    if (s1[i] == s2[i]) {
        s+=s1[i];
    }
    else {
        break;
    }
}
s.reverse();
cout<<s;

- harrypotter0 September 16, 2017 | Flag Reply
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0
of 0 vote

//C++ Code 
for( int i = a.length()-1, j = b.length()-1; i >= 0 && j >= 0 && a[i] == b[j]; i--;j--);
if ( i < a.length()-1 ) return a.substr(i);
else
return "";

- harrypotter0 September 16, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

std::string findCommonSuffix(std::string s1, std::string s2)
{
    int i, j; 
    for(i = s1.size()-1, j = s2.size()-1 ; i >= 0 && j >= 0; --i, --j)
    {
        if(s1[i] != s2[j]) break;
        if(i == 0 || j == 0) break;
    }
    std::string result;
    if(i == 0)
         result = s1;
    else if(j == 0)
        result = s2;
    else
        result = s1.substr(i+1, s1.size() - i);
    return result;
}

- steephen September 16, 2017 | Flag Reply
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0
of 0 vote

String s1 = "cornfiled";
String s2 = "Exfiled";
String suffix;

if ( s1.length() > s2.length()){
suffix = searchSuffix(s1, s2);
}else{
suffix = searchSuffix(s2, s1);
}

System.out.println("Matching suffix is :" + suffix);
}

static String searchSuffix(String searchString, String searchInString){

String sbSearch = null;
for (int i=0; i<searchString.length(); i++){
sbSearch = searchString.substring(i);
System.out.println("sbSearch :" + sbSearch);
if(searchInString.contains(sbSearch)){
return sbSearch;
}
}

return null;
}

- Chinmay October 04, 2017 | Flag Reply
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0
of 0 vote

public static void main(String[] args){

String s1 = "cornfiled";
String s2 = "Exfiled";
String suffix;

if ( s1.length() > s2.length()){
suffix = searchSuffix(s1, s2);
}else{
suffix = searchSuffix(s2, s1);
}

System.out.println("Matching suffix is :" + suffix);
}

static String searchSuffix(String searchString, String searchInString){

String sbSearch = null;
for (int i=0; i<searchString.length(); i++){
sbSearch = searchString.substring(i);
System.out.println("sbSearch :" + sbSearch);
if(searchInString.contains(sbSearch)){
return sbSearch;
}
}

return null;
}

- Chinmay October 04, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

String[] words = {"cornfiled", "Exfiled", "eled"};
        Stream.of(words)
                .sorted(Comparator.reverseOrder())
                .reduce((r, l) ->
                {
                    String suff = "";
                    for (int i = r.length() - 2; i >= 0; i--) {
                        String temp = r.substring(i);
                        if (!l.endsWith(temp)) {
                            break;
                        }
                        suff = temp;
                    }
                    return suff;

                }).ifPresent(System.out::println);

- Eugeniy October 28, 2017 | Flag Reply
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0
of 0 vote

String[] words = {"cornfiled", "Exfiled", "eled"};
        Stream.of(words)
                .sorted(Comparator.reverseOrder())
                .reduce((r, l) ->
                {
                    String suff = "";
                    for (int i = r.length() - 2; i >= 0; i--) {
                        String temp = r.substring(i);
                        if (!l.endsWith(temp)) {
                            break;
                        }
                        suff = temp;
                    }
                    return suff;

                }).ifPresent(System.out::println);

- Eugeniy October 28, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
	        int temp;
	        try{
	        System.out.println("enter string 1:");
	        String str1=sc.nextLine();
	        System.out.println("enter string 2:");
	        String str2=sc.nextLine();
	        if(str1.length()>str2.length())
	        {
	        	if(str1.contains(str2)){
	        		temp=str1.indexOf(str2);
	        		System.out.print(str1.substring(temp));
	        	}
	        }
	        else
	        {
	        	if(str2.contains(str1)){
	        		temp=str2.indexOf(str1);
	        		System.out.print(str2.substring(temp));
	        	}
	        }
	        }
	        catch(Exception e){
	        	System.err.print(e);
	        }
	        
	}

- shrikantjoshi1991 November 02, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def commonSufix(str1, str2):

    i1 = len(str1)-1
    i2 = len(str2)-1

    while i1 >= 0 and i2 >= 0 and str1[i1] == str2[i2]:
        i1 -= 1
        i2 -= 1
    return str1[i1+1:]

- desaivaibhav94 October 14, 2018 | Flag Reply
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0
of 0 vote

def longest_common_suffix(arr)
  suffix = arr[0]
  i = 0

  while i < arr.length
    temp = 0
    until suffix.nil? || arr[i].end_with?(suffix)
      temp += 1
      suffix = suffix[temp, suffix.length]
    end
    
    i += 1
  end

  suffix
end

input = ['bakersfield', 'intensefield', 'vfield']
puts longest_common_suffix(input)

- Anonymous December 30, 2018 | Flag Reply


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