// C Program to Find second most frequent character in a String
#include <stdio.h>
#include <string.h>
int main()
{
char str[110],answer;
int i, length;
int max = -1;
int frequency[256] = {0};
printf("\nEnter the String of characters: ");
gets(str);
length = strlen(str);
for(i = 0; i < length; i++)
{
frequency[str[i]]++;
}
for(i = 0; i < length; i++)
{
if(max < frequency[str[i]])
{
max = freq[str[i]];
answer = str[i];
}
}
printf("\nThe second most frequent character in a Given String is= %c ", answer);
return 0;
}
Output
Enter the String of characters: aabbbcde
The second most frequent character in a Given String is= a
// Java Program to find the second most frequent character in a given string
import java.util.*;
public class Main
{
static final int CHARS = 256;
static char secondMostFreqChar(String str1)
{
int[] ch = new int[CHARS];
for (int i = 0; i < str1.length (); i++)
(ch[str1.charAt (i)])++;
int ch_first = 0, ch_second = 0;
for (int i = 0; i < CHARS; i++)
{
if (ch[i] > ch[ch_first])
{
ch_second = ch_first;
ch_first = i;
}
else if (ch[i] > ch[ch_second] && ch[i] != ch[ch_first])
ch_second = i;
}
return (char) ch_second;
}
public static void main (String args[])
{
Scanner sc=new Scanner(System.in);
System.out.println ("Enter a string: ");
String str1=sc.next();
System.out.println ("The given string is: " + str1);
char res = secondMostFreqChar (str1);
if (res != '\0')
System.out.println ("The second most frequent character in the string is: " +res);
else
System.out.println ("No second most frequent character found in the string");
}
}
Output
Enter a string: visit prepinsta
The given string is: visit prepinsta
The second most frequent character in the string is: p
All Platforms
Programming
Aptitude
Syllabus
Off Campus
0
a = input()
b = dict()
c = []
for i in a:
b[i] = 0
for i in b:
x = a.count(i)
b[i] = x
if x not in c:
c.append(x)
c.sort(reverse=True)
if len(c) == 1:
exit(print(“No Second most frequent character”))
for i in b:
if b[i] == c[1]:
print(“Second most frequent character is”, i)
#python
from collections import Counter
string = input()
counter_string = Counter(string)
max_two = counter_string.most_common(2)
if max_two[1][1] <= 1:
print("No Second most frequent character")
else:
print(max_two[1][0], max_two[1][1])
// C Program to Find second most frequent character in a String
#include
#include
int main()
{
char s[1000];
int max=-1,smax=-1,i,freq[1000]={0};
scanf(“%s”,s);
for(i=0;i<strlen(s);i++){
freq[s[i]]=freq[s[i]]+1;
}
for(i=0;imax){
smax=max;
max=i;
}
else if(freq[s[i]] > freq[s[smax]] && freq[s[i]] != freq[s[max]])
smax = i;
}
printf(“%c-%d\n”,s[smax],freq[s[smax]]);
printf(“%d-%d\n”,max,smax);
return 0;
}
JAVA:(first most and second most):
import java.util.*;
public class Main
{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String str = sc.nextLine(); //”grass is greener on the other side”
int[] freq = new int[str.length()];
char minChar = str.charAt(0), maxChar = str.charAt(0);
int i, j, min, max;
char minChar1 = str.charAt(0), maxChar1 = str.charAt(0);
int min1,max1;
//Converts given string into character array
char string[] = str.toCharArray();
for(int k = 0; k < string.length; k++){
System.out.print(string[k]+" ");
}
//Count each word in given string and store in array freq
for(i = 0; i < string.length; i++) {
freq[i] = 1;
for(j = i+1; j < string.length; j++) {
if(string[i] == string[j] && string[i] != ' ' && string[i] != '0') {
freq[i]++;
//Set string[j] to 0 to avoid printing visited character
string[j] = '0';
}
}
}
System.out.print("\n");
for(int l = 0; l < string.length; l++){
System.out.print(string[l]+" ");
}
System.out.print("\n");
for(int m = 0; m < string.length; m++){
System.out.print(freq[m]+" ");
}
System.out.print("\n");
//Determine minimum and maximum occurring characters
min = max = freq[0];
for(i = 0; i freq[i] && string[i] != ‘0’) {
min = freq[i];
minChar = string[i];
}
//If max is less than frequency of a character
//then, store frequency in max and corresponding character in maxChar
if(max < freq[i]) {
max = freq[i];
maxChar = string[i];
}
}
System.out.println(min);
System.out.println(max);
min1 = max1 = freq[0];
for(int z = 0; z freq[z] && string[z] != ‘0’ && freq[z] != min ) {
min1 = freq[z];
minChar1 = string[z];
}
//If max is less than frequency of a character
//then, store frequency in max and corresponding character in maxChar
if(max1 < freq[z] && freq[z] != max ) {
max1 = freq[z];
maxChar1 = string[z];
}
}
System.out.println(min1);
System.out.println(max1);
System.out.println("Minimum occurring character: " + minChar);
System.out.println("Maximum occurring character: " + maxChar);
System.out.println("2nd most Minimum occurring character: " + minChar1);
System.out.println("2nd most Maximum occurring character: " + maxChar1);
}
}
#include
#include
#include
using namespace std;
int main()
{
string s; int val = 0, max, sec, a[26] = {0}, b[26] = {0};
cin >> s;
sort(s.begin(), s.end());
for(auto i:s)
{
val = (int)tolower(i) – 97;
a[val]++;
b[val]++;
}
sort(a, a+26);
max = a[25];
int val1 = 1, i = 24;
while (val1 == 1)
{
if(max != a[i])
{
sec = a[i];
val1 = 0;
}
i–;
}
int pos = 0;
for(int i =0; i < 26; i++)
{
if ( b[i] == sec )
{
pos = i + 97 ;
break;
}
}
cout <<"\n ans is " << (char)pos ;
return 0;
}
str = “aabababaccdddeffffff”
dict={}
for ele in str:
if(ele in dict.keys()):
dict[ele]+=1
else:
dict[ele]=1
print(dict)
l=list(dict.values())
for j in range (2):
for i in range (len(l)-1):
if(l[i]>l[i+1]):
l[i],l[i+1]=l[i+1],l[i]
c=l[::-1][1]
if(c!=1):
print(c)
else:
print(‘Not Possible’)
int main()
{
string s ;
int ans = 0;
int poi =0;
getline(cin,s);
transform(s.begin(), s.end(), s.begin(), ::tolower);
int freq[26] = {0};
for(int i =0 ;i<s.length();i++)
{
freq[s[i] – 'a']++;
}
for(int i =0 ;i<26;i++)
{
if(ans<freq[i])
{
ans = freq[i];
poi = i;
}
}
freq[poi] = freq[poi] – ans;
ans =0;
for(int i =0 ;i<26;i++)
{
if(ans<freq[i])
{
ans = freq[i];
poi = i;
}
}
char chara = poi + 'a';
cout<<"The second most frequent character in a Given String is "<<chara;
return 0;
}
n = input(“enter : “)
s = {i for i in n}
m = {}
for i in s:
m.update({i:n.count(i)})
a = m.pop(max(m))
b = max(m)
if a >= 2:
print(“No”)
else:
print(b)
Python Solution
a=’aabababa’
b=list(a)
c=sorted(set(b),key=b.count,reverse=True)
print(c[1])
#include
#include
#include
using namespace std;
bool sortByVal(pair p,pair q)
{
return p.second > q.second;
}
char secondMostFrequent(string s)
{
unordered_map map;
for(char c : s)
{
if(map.find(c) == map.end())
map[c] = 1;
else
map[c] +=1;
}
vector<pair> freqs;
for(auto kv : map)
freqs.push_back(kv);
sort(freqs.begin(),freqs.end(),sortByVal);
return freqs[1].first;
}
int main() {
string s;
cin >> s;
char output = secondMostFrequent(s);
cout << output << endl;
return 0;
}
#python
b=str(input(“input”))
a=b.replace(” “,””)
c={}
for i in a:
if i in c:
c[i]+=1
else:
c[i]=1
s=sorted(c.values())
g=s[-2]
print(list(c.keys())[list(c.values()).index(g)])
#Using Counter
from collections import Counter
s = ‘aababc’
a = Counter(s)
g = sorted(a.values())[-2]
print(list(a.keys())[list(a.values()).index(g)])
#python
def second_most_frequent_char(str):
no_of_char=256
count=[0]*no_of_char
for i in range(len(str)):
count[ord(str[i])]+=1
first,second=0,0
for i in range(no_of_char):
if(count[i]>count[first]):
second=first
first=i
elif(count[i]>count[second] and count[i]!=count[first]):
second=i
return chr(second)
string=input()
res=second_most_frequent_char(string)
if(res!=’\0′):
print(“second most frequent character is:”,res)
else:
print(‘no second most frequent char’)