N string values are passed as input to the program. Each string will contain only the alphabets a-z in lower case. A given alphabet may be repeated any number of times. The program must print the count C of the alphabets that are present in all the N string values.
Input Format:
The first line contains N.
Next N lines contain the N string values.
Output Format:
The first line contains C.
Boundary Conditions:
2 <N <500
1 <Length of the string value <1000
Example Input/Output 1:
Input:
3
mnppqqr
ajkmnnm
poormanagement
Output:
2
Explanation :
Only 2 alphabets m and n are present in all the three string values.
#include <iostream>
using namespace std;
int main(int argc, char** argv)
{
int n;
cin >> n;
string a[n],temp;
for(int i=0;i<n;i++)
{
cin >> a[i];
}
int b[n][26]={0},c;
for(int i=0;i<n;i++)
{
temp = a[i];
for(int j=0;j<temp.length();j++)
{
c = temp[j]-97;
b[i][c]++;
}
}
int flag,count = 0;
for(int i=0;i<26;i++)
{
flag = 1;
for(int j=0;j<n;j++)
{
if(b[j][i] <= 0)
{
flag = 0;
}
}
if(flag == 1)
{
count++;
}
}
cout << count;
return 0;
}
n = int(input()) # Read the number of strings (N).
letter_count = [0] * 26 # Initialize an array to count occurrences of each letter.
for _ in range(n):
s = input() # Read each string.
# Create a set to track which letters are present in the current string.
present = set()
# Mark each letter as present in the current string.
for c in s:
present.add(c)
# Update the letter_count list based on the current string.
for i in range(26):
if chr(ord('a') + i) in present:
letter_count[i] += 1
count = 0 # Initialize the count of letters present in all strings.
# Count the letters that are present in all strings.
for i in range(26):
if letter_count[i] == n:
count += 1
print(count) # Output the count.
#include <stdio.h>
int main() {
int n;
scanf("%d", &n); // Read the number of strings (N).
int letterCount[26] = {0}; // Initialize an array to count occurrences of each letter.
for (int i = 0; i < n; i++) {
char s[1000];
scanf("%s", s); // Read each string.
// Create an array to track which letters are present in the current string.
int present[26] = {0};
// Mark each letter as present in the current string.
for (int j = 0; s[j] != ' '; j++) {
present[s[j] - 'a'] = 1;
}
// Update the letterCount array based on the current string.
for (int j = 0; j < 26; j++) {
if (present[j]) {
letterCount[j]++;
}
}
}
int count = 0; // Initialize the count of letters present in all strings.
// Count the letters that are present in all strings.
for (int i = 0; i < 26; i++) {
if (letterCount[i] == n) {
count++;
}
}
printf("%dn", count); // Output the count.
return 0;
}
using System;
class Program {
static void Main() {
int n = int.Parse(Console.ReadLine()); // Read the number of strings (N).
int[] letterCount = new int[26]; // Initialize an array to count occurrences of each letter.
for (int i = 0; i < n; i++) {
string s = Console.ReadLine(); // Read each string.
// Create an array to track which letters are present in the current string.
bool[] present = new bool[26];
// Mark each letter as present in the current string.
foreach (char c in s) {
present[c - 'a'] = true;
}
// Update the letterCount array based on the current string.
for (int j = 0; j < 26; j++) {
if (present[j]) {
letterCount[j]++;
}
}
}
int count = 0; // Initialize the count of letters present in all strings.
// Count the letters that are present in all strings.
for (int i = 0; i < 26; i++) {
if (letterCount[i] == n) {
count++;
}
}
Console.WriteLine(count); // Output the count.
}
}
#include <iostream>
using namespace std;
int main() {
int n;
cin >> n; // Read the number of strings (N).
int letterCount[26] = {0}; // Initialize an array to count occurrences of each letter.
for (int i = 0; i < n; i++) {
string s;
cin >> s; // Read each string.
// Create an array to track which letters are present in the current string.
bool present[26] = {false};
// Mark each letter as present in the current string.
for (char c : s) {
present[c - 'a'] = true;
}
// Update the letterCount array based on the current string.
for (int j = 0; j < 26; j++) {
if (present[j]) {
letterCount[j]++;
}
}
}
int count = 0; // Initialize the count of letters present in all strings.
// Count the letters that are present in all strings.
for (int i = 0; i < 26; i++) {
if (letterCount[i] == n) {
count++;
}
}
cout << count << endl; // Output the count.
return 0;
}
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