Problems

maximum length of palindrome can be formed?

Example1:

Input:

3

malayalam

ere

level

Output:

15

Example2:

Input:

4

a

b

cd

ab

Output:

5

C-code

compiler

#include<stdio.h>

int main() {

    int n;

    scanf("%d",&n);

    int fre1[26]={0};

    for(int i=0;i<n;i++)

        char str[100];

        scanf("%s",str);

        for(int j=0;str[j]!='\0';j++)

            fre1[str[j]-'a']++;

    int value=0,flag=0;

    for(int i=0;i<26;i++)

        if(fre1[i]%2)

        flag=1;

        if(fre1[i]>=2)

        value=value+(2*(fre1[i]/2));

    printf("%d",flag?value+1:value);

Braces -valid or not?

Example1:

Input:

[{}]

Output:

Valid

Example2:

Input:

{[[])]

Output:

Invalid

C-code

compiler

#include<stdio.h>

int main() {

   char str[1001];

   int len;

   scanf("%s%n",str,&len);

   int flag=0;

   if(len%2)

       printf("Invalid");

       return 0;

   else

   for(int i=0;i<len/2;i++)

       if(str[i]=='{'&&str[len-i-1]!='}')

       flag=1;

       else if(str[i]=='('&&str[len-i-1]!=')')

       flag=1;

       else if(str[i]=='['&&str[len-i-1]!=']')

       flag=1;

  printf(flag?"Invalid":"Valid");

Expand alphabets

#include<stdio.h>

#include<stdlib.h>

int main() {

  char str[100],tempstr[100];

  int len;

  scanf("%s%n",str,&len);

  int i=0,sindex=0;

  while(i<len)

      if(isalpha(str[i])&&isdigit(str[i+1]))

          int j=i+1,num=0;

          while(isdigit(str[j]))

              num=num*10+(str[j]-'0');

              j++;

          for(int index=0;index<num;index++)

             tempstr[sindex++]=str[i];

          i=j;

      else

         tempstr[sindex++]=str[i];

          i++;

  tempstr[sindex]='\0';

  printf("%s",tempstr);

XO-Game

Example

Inputs1:

X X X T

. . . .

O O . .

. . . .

Output1:

X won

Inputs2:

X O X T

X X O O

O X O X

X X O O

Output2:

draw

Inputs3:

X O X .

O X . .

. . . .

Output3:

Game not completed

Inputs4:

O O X X

O X X X

O X . T

O . . O

Output4:

O won

Inputs5:

X X X O

. . O .

. O . .

T . . .

Output5:

O won

Inputs6:

O X X X

X O . .

. . O .

. . . O

Output6:

O won

C-Code

compiler

#include<stdio.h>

int main() {

    char mat[4][4];

    int flagx=0,flago=0,t=0;

    for(int row=0;row<4;row++)

    {   int countx=0,counto=0,countt=0;

        for(int col=0;col<4;col++)

        { char str[100];

            scanf("%s",str);

            mat[row][col]=str[0];

            if(mat[row][col]=='X')

            countx++;

            else if(mat[row][col]=='O')

            counto++;

            else if(mat[row][col]=='T')

            countt++;

            t=1;

        if(countx==4||(countx==3&&countt==1))

            printf("X is won");

            return 0;

        else if(counto==4||(counto==3&&countt==1))

            printf("O is won");

            return 0;

    for(int col=0;col<4;col++)

    {   int countx=0,counto=0,countt=0;

        for(int row=0;row<4;row++)

            if(mat[row][col]=='X')

            countx++;

            else if(mat[row][col]=='O')

            counto++;

            else if(mat[row][col]=='T')

            countt++;

            t=1;

        if(countx==4||(countx==3&&countt==1))

            printf("X is won");

            return 0;

        else if(counto==4||(counto==3&&countt==1))

            printf("O is won");

            return 0;

        int i=0;

         int countx=0,counto=0,countt=0;

        while(i<4)

           if(mat[i][i]=='X')

            countx++;

            else if(mat[i][i]=='O')

            counto++;

            else if(mat[i][i]=='T')

            countt++;

            t=1;

            i++;

        if(countx==4||(countx==3&&countt==1))

            printf("X is won");

            return 0;

        else if(counto==4||(counto==3&&countt==1))

            printf("O is won");

            return 0;

        i=0;

        int j=3,countO=0,countX=0,countT=0;

        while(i<=3)

           if(mat[i][j]=='X')

            countX++;

            else if(mat[i][j]=='O')

            countO++;

            else if(mat[i][j]=='T')

            countT++;

            t=1;

            i++;

            j--;

        if(countX==4||(countX==3&&countT==1))

            printf("X is won");

            return 0;

        else if(countO==4||(countO==3&&countT==1))

            printf("O is won");

            return 0;

        if(!t)

        printf("Game is incomplete");

        return 0;

        printf("Draw");

Common word in a two string last and first

Example:

Input:

mayday

daybreak

Output:

day

Example:

Input:

kgear

gearr

Output:

gear

C-Code

compiler

#include<stdio.h>

#include<string.h>

int main() {

    char str1[1001],str2[1001];

    scanf("%s %s",str1,str2);

    int len=strlen(str1);

    int i=0,j=0;

    while(i<len)

        if(str1[i]==str2[j])

            i++;

            j++;

        else

            i++;

            j=0;

    str2[j]='\0';

    printf("%s",str2);

Series Team Score

Two soccer teams A and B play a series of matches over a period of time. In a match, the winning team gets 3 points. If the match ends in a tie (draw) with both teams scoring same goals, then both the teams get one point each. The losing team does not get any point. The program must accept the goals scored by both team A and B in certain number of matches and print the cumulative scores of team A and B separated by a space.

Example Input/Output 1:

Input:

3 5 1 3 2 0

Output:

7 1

Explanation: Team A drew the first match and hence both team A and B got one point each. Team A won both matches two and three and hence got additional 6 points. So the final score of team A is 7 and team B is 1.

C-code

Compiler

#include<stdio.h>

int main() {

    int arr[1001],index=-1;

    while(scanf("%d",&arr[++index])>0);

    int start=0,end=index/2,team1=0,team2=0;

    while(start<index/2)

        if(arr[start]>arr[end])

            team1+=3;

        else if(arr[start]==arr[end])

                team1+=1;

                team2+=1;

        else if(arr[start]<arr[end])

            team2+=3;

        start++;

        end++;

    printf("%d %d",team1,team2);

String-Count articles

A string S is passed as the input. The program must print the number of articles in S. The string S passed as the input NEED NOT be correct grammatically.

Example Input/Output 1:

Input:

I went to a movie yesterday along with an old man.

Output:

2

Example Input/Output 2:

Input:

In 1885 it was banished from the shelves of the Concord Public Library, an act that attracted a lot of publicity and discussion in the press. It is still frequently in the news, as various schools and school systems across the country either ban it from or restore it to their classrooms. The texts and illustrations below attempt to capture both the novel's achievement and some aspects of its controversiality.

Output:

8

Example Input/Output 3:

Input:

Count the articles which are a, an and the.

Output:

4

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int isarticle(char str[])

    int len=strlen(str);

    if(len==1&&tolower(str[0])=='a')

        return 1;

    else if(len==2&&tolower(str[0])=='a'&&str[1]=='n')

        return 1;

    else if(len==3&&tolower(str[0])=='t'&&str[1]=='h'&&str[2]=='e')

        return 1;

    else

        return 0;

int main()

char str[100];

int count=0;

while(scanf("%s",str)>0)

{   int len=strlen(str);

    if(str[len-1]=='.'||str[len-1]==',')

        str[len-1]=NULL;

    if(isarticle(str))

        count++;

printf("%d",count);

Message-Encryption

To encrypt messages Jil will first decide on the number of columns C to use. Then Jil will pad the message with letters chosen randomly so that they form a rectangular matrix. Finally Jil will write down the message navigating the rows from left to right and then from right to left. The program must accept the encrypted message M as input and then extract and print the original message (along with any additional padding letters) from the encrypted one based on the value of C.

Sample Input/Output:

Example 1:

Input:

midinadiazne

3

Output:

madeinindiaz

Explanation:

m i d

a n i

d i a

e n z

Here z is the padding letter. The navigating across the rows mid (left to right) ina (right to left) and so on we come up with the encrypted message midinadiazne.

Example 2:

Input:

loaesfbnaiordilertenrdhdw

5

Output:

lionroaredandthebirdsflew

Explanation:

l o a e s

i a n b f

o r d i l

n e t r e

r d h d w

Here there are no padding letters. The navigating across the rows left to right and then from right to left we get loaesfbnaiordilertenrdhdw

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

char str[1001],msge[100][100];

int n,index=-1;

scanf("%s\n%d",str,&n);

for(int i=0;i<strlen(str)/n;i++)

    for(int j=0;j<n;j++)

        if(i%2==0)

            msge[i][j]=str[++index];

        else

            msge[i][j]=str[index--];

    index=index+n;

for(int col=0;col<n;col++)

    for(int row=0;row<strlen(str)/n;row++)

        printf("%c",msge[row][col]);

Maximum sum(prizemoney)-Time complexity

An array of N integers (both positive and negative) is given as the input to the program.

The program must print the maximum sum of the subarray in the given array as the output.

Boundary Condition(s): 1 <= N <= 1000

Input Format:

The first line contains N.

The second line contains N integers separated by space.

Output Format:

The first line contains the maximum subarray sum.

Example Input/Output 1:

Input:

7 2 1 -6 4 -2 8 -2

Output:

10

Explanation:

The subarray 4 -2 8 has the maximum sum 10.

Example Input/Output 2:

Input:

10 73 64 18 -18 44 -4 -17 -28 -59 54

Output:

181

C-code

Compiler

#include <stdio.h>

int main() {

int N;

scanf("%d",&N);

int num,sum=0,max=0;

for(int index=0;index<N;index++)

scanf("%d",&num);

sum=sum+num;

if(sum<0)

sum=0;

else if(sum>max)

max=sum;

printf("%d",max);

Alphabet printing

Example Input/Output 1:

Input:

i a z

Output:

i h g f e d c b a

a b c d e f g h i j k l m n o p q r s t u v w x y z

i j k l m n o p q r s t u v w x y z

Example Input/Output 2:

Input:

a j p

Output:

a b c d e f g h i j

j k l m n o p

a b c d e f g h i j k l m n o p

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

void print(char start,char end)

    if(start=='z'&&end=='z')

        printf("z\n");

        return;

    if(start=='z')

    printf("z ");

    start='a';

   if(start<end)

    while(start<=end)

    printf("%c ",start++);

   else

       while(start>=end)

    printf("%c ",start--);

    printf("\n");

int main()

char ch1,ch2,ch3;

scanf("%c %c %c",&ch1,&ch2,&ch3);

print(ch1,ch2);

print(ch2,ch3);

print(ch1,ch3);

Frequency of a word

C-code

Compiler

#include<stdio.h>

#include<string.h>

int main() {

    char a[100];

    int fresmall[26]={0};

    int frecap[26]={0};

    scanf("%s",a);

    for(int i=0;i<strlen(a);i++)

        if(a[i]>='a'&&a[i]<='z')

            fresmall[a[i]-'a']++;

        else if(a[i]>='A'&&a[i]<='Z')

            frecap[a[i]-'A']++;

    printf("FREQUENCY OF SMALL LETTER IN A GIVEN WORD:\n");

    for(int i=0;i<26;i++)

        if(fresmall[i]>0)

            printf("%c->%d\n",i+'a',fresmall[i]);

 printf("\nFREQUENCY OF CAPTIAL LETTER IN A GIVENWORD:\n");

     for(int i=0;i<26;i++)

        if(frecap[i]>0)

            printf("%c->%d\n",i+'A',frecap[i]);

    printf("\nFREQUENCY IN THE ORDER OF GIVEN INPUT:\n");

    for(int i=0;i<strlen(a);i++)

        for(int j=0;j<26;j++)

            if(a[i]==j+'a'&&fresmall[j]!=0)

                printf("%c->%d\n",a[i],fresmall[j]);

                fresmall[j]=0;

            else

            if(a[i]==j+'A'&&frecap[j]!=0)

                printf("%c->%d\n",a[i],frecap[j]);

                frecap[j]=0;

Pangram

Note:A line or word which contains all the alphabetic letter is known as pangram

C-code

Compiler

#include<stdio.h>

int main() {

    char a[100];

    int fre[26]={0};

    int total=0;

    scanf("%[^\n]",a);

    for(int i=0;i<strlen(a);i++)

        if(a[i]>='a'&&a[i]<='z')

            total=total+!fre[a[i]-'a'];

            fre[a[i]-'a']=1;

    if(total==26)

        printf("Given is pangram");

    else

        printf("not a pangram");

No adjacent character are same

Example:

Input

aaabc

Output

abaca

C-code

Compiler

#include<stdio.h>

int main() {

    char str[100];

    int len,k=0,flag=0;

    scanf("%s%n",str,&len);

    while(k<len)

        for(int i=0;i<len;i++)

        {   if(i==len-1)

            continue;

            else if(str[i]==str[i+1])

                flag=1;

                break;

        if(flag==0)

            printf("%s",str);

            return 0;

        else

            for(int i=1;i<len;i++)

            {   if(i==len-1)

                continue;

                else if(str[i-1]==str[i]&&str[i]!=str[i+1])

                    char temp=str[i];

                    str[i]=str[i+1];

                    str[i+1]=temp;

        flag=0;

        k++;

    printf("Not possible");

minimum to make as palindrome

Example:

Input

mddam

Output

0 //if a string has odd length it can have one odd character

Example:

Input

aabbccasd

Output

2

C-code

Compiler

#include<stdio.h>

#include<string.h>

int main() {

    char str[100];

    int fre[26]={0},len,count=0;

    scanf("%s%n",str,&len);

    for(int i=0;i<strlen(str);i++)

        if(str[i]>='a'&&str[i]<='z')

            fre[str[i]-'a']++;

    for(int i=0;i<26;i++)

        if(fre[i]!=0)

            count=count+(fre[i]%2);

    if(len%2==1)//for odd sequence

        printf("%d",count-1);

    else//for even sequence

        printf("%d",count);

print consecutive digit in reverse order

C-code

Compiler

#include<stdio.h>

#include<string.h>

int arr[9],count=0;

void check(int num)

    if(!arr[num])

        printf("%d",num);

        arr[num]=1;

        count=1;

int main() {

   char str[100];

   scanf("%s",str);

   for(int index=strlen(str)-1;index>0;index--)

       if(str[index]==str[index-1])

            check((str[index]-'0'));

   if(!count)

        printf("-1");

Print maximum length of string in an line

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

#include<string.h>

int main()

char str[100],max[100],len=0;

while(scanf("%s",str)>0)

    int slen=strlen(str);

    if(slen>len)

        len=slen;

        strcpy(max,str);

printf("%s",max);

Remove common character in string

Example:

Input

skillrack

race

Output:

skillk

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

#include<string.h>

int main()

char a[1001],b[100];

scanf("%s%s",a,b);

char temp[100];

int k=0;

for(int i=0;i<strlen(b);i++)

    for(int j=0;j<strlen(a);j++)

        if(b[i]==a[j])

            a[j]='0';

for(int i=0;i<strlen(a);i++)

    if(a[i]!='0')

        temp[k++]=a[i];

temp[k]='\0';

printf("%s",temp);

Weight calculate

C-code

Compiler

#include<stdio.h>

int main() {

    int n;

    scanf("%d\n",&n);

    int num[n],wei[n];

    for(int i=0;i<n;i++)

    {   int weigth=0;

        scanf("%d\n",&num[i]);

        if(num[i]%4==0&&num[i]%6==0)

            weigth=weigth+4;

        if(num[i]%2==0)

            weigth=weigth+3;

        for(int j=2;j<=num[i]/2;j++)

            if(num[i]==(j*j))

                weigth=weigth+5;

        wei[i]=weigth;

    for(int i=0;i<n;i++)

        for(int j=i+1;j<n;j++)

            if(wei[i]>wei[j])

                int temp=wei[i];

                wei[i]=wei[j];

                wei[j]=temp;

                temp=num[i];

                num[i]=num[j];

                num[j]=temp;

            else if(wei[i]==wei[j])

                if(num[i]>num[j])

                    int temp=wei[i];

                wei[i]=wei[j];

                wei[j]=temp;

                temp=num[i];

                num[i]=num[j];

                num[j]=temp;

    for(int i=n-1;i>=0;i--)

        printf("%d\n",num[i]);

2d array(half test case passed)

C-code

Compiler

#include<stdio.h>

#include<string.h>

int main() {

   char a[1001];

   char check[100];

   int size,k=0;

   scanf("%s\n%s\n%d",a,check,&size);

   char matrix[size][size];

   for(int i=0;i<size;i++)

   {    if(a[k]=='\0')

       break;

       for(int j=0;j<size;j++)

           matrix[i][j]=a[k];

        k++;

    for(int i=0;i<size;i++)

   {    int flag1=0,start1=0;

       for(int j=0;j<size;j++)

          if(matrix[i][j]==check[flag1])

              if(flag1==0)

                  start1=j;

              flag1++;

       if(flag1==strlen(check))

           printf("%d.%d %d.%d",i,start1,i,start1+(strlen(check)-1));

           return 0;

    for(int i=0;i<size;i++)

   {    int flag2=0,start2=0,colum;

       for(int j=0;j<size;j++)

           if(matrix[j][i]==check[flag2])

              if(flag2==0)

                  start2=j;

                  colum=i;

              flag2++;

    if(flag2==strlen(check))

           printf("%d.%d %d.%d",start2,colum,start2+(strlen(check)-1),colum);

           return 0;

Symmetric pair (using c)

Example:

Input:

(1,3) (3,1) (1,2) (1,2)

Output:

(1,3) (3,1)

C-code

Compiler

#include<stdio.h>

#include <stdlib.h>

int main()

int x,y,a[101],i=0;

while(scanf("(%d,%d) ",&x,&y)>0)

    a[i++]=x;

    a[i++]=y;

for(int j=1;j<i;j++)

    for(int k=j+1;k<i;k++)

    {   if(j==i-1||k==i-1)

        continue;

        else if(a[j-1]==a[k+1]&&a[j]==a[k])

            printf("(%d,%d) (%d,%d)",a[j-1],a[j],a[k],a[k+1]);

            printf("\n");

            break;

Symmetric pair(using java method 1)

import java.util.*;

class Check

{   public int x,y;

    Check(int x,int y)

        this.x=x;

        this.y=y;

public class Binary{

    Set<String> set1=new LinkedHashSet<String>();

    public  void find(Check c)

            String p="{"+c.x+","+c.y+"}";

            set1.add(p);

            String rev="{"+c.y+","+c.x+"}";

            if(set1.contains(rev))

                System.out.println(rev+"|"+p);

                System.out.println();

    public static void main (String[] args) {

        Scanner sc=new Scanner(System.in);

        Binary b=new Binary();

        while(sc.hasNext())

           int x=sc.nextInt();

           int  y=sc.nextInt();

            Check c=new Check(x,y);

            b.find(c);

Symmetric pair(using java method 2)

import java.util.*;

class Check

{   public int x,y;

    Check(int x,int y)

        this.x=x;

        this.y=y;

public class Binary{

   static  Set<String> set1=new LinkedHashSet<String>();

     static int count=0;

    public static void find(Check []c)

    {   for(Check m:c)

            String p="{"+m.x+","+m.y+"}";

            set1.add(p);

            String rev="{"+m.y+","+m.x+"}";

            if(set1.contains(rev))

                System.out.println(p+" | "+rev);

                System.out.println();

                count=1;

    public static void  main (String[] a) {

        Scanner sc=new Scanner(System.in);

        Binary b=new Binary();

        System.out.println("Enter the number of pairs");

        int num=sc.nextInt();

        Check [] c=new Check[num];

        if(num<=1)

        System.out.println("Invalid Input");

        System.exit(0);

        System.out.println("Symmetric Pairs:");

        for(int i=0;i<num;i++)

            int x=sc.nextInt();

            int y=sc.nextInt();

            c[i]=new Check(x,y);

        find(c);

        if(count==0)

            System.out.println("No Symmetric pairs found");

Given string is present or Not?

Maximum sum(prize money)-Time complexity

A string S and N words are given as the input to the program. The program must print Yes if each word among the N words can be formed using the alphabets in the string S. Else the program must print No as the output.

Boundary Condition(s): 1 <= Length of each string <= 1000 1 <= N <= 100

Input Format: The first line contains string S. The second line contains N. The next N lines contain N string values one in each line.

Output Format: The first line contains Yes or No based on the given conditions.

Example Input/Output 1:

Input:

telephonegrammary

3

memory

elegant

Output:

Yes

Example Input/Output 2:

Input:

amsterdamacademy

2

amy

racer

Output:

No Explanation: racer has two r which are not available in the string S.

C-code

Compiler

#include <stdio.h>

#include<string.h>

 int main()

    int N,strfre[26]={0};

    char str[1001];

    scanf("%s\n",str);

    scanf("%d\n",&N);

    for(int i=0;i<strlen(str);i++)

        strfre[str[i]-'a']++;

    for(int i=0;i<N;i++)

        char temp[100];

        int tempfre[26]={0};

        scanf("%s\n",temp);

        for(int k=0;k<strlen(temp);k++)

            tempfre[temp[k]-'a']++;

        for(int d=0;d<26;d++)

            if(tempfre[d]!=0&&tempfre[d]>strfre[d])

                printf("NO");

                return 0;

    printf("YES");