Given two strings, S1 and S2 and an integer K.A string str can be made by performing the following operations:
- Take an Empty String Initially
- Append S1 one times
- Append S2 two times
- Append S1 three times
- Append S2 four times
- Append S1 five times
- Append S2 six times
- You can keep building this string until it reaches K
The task is to find the character at the kth position in the new string str.
Input: S1 = “a”, S2 = “bc”, k = 4
Output: b
Explanation: s3 = “a” + “bcbc” +”aaa” + “bcbcbcbc” ==> abcbcaaabcbcbcbc. so the character on the 4th index is ‘b’Input: S1 = “Geeks”, S2 = “for”, k = 3
Output: e
Explanation: s3 = “Geeks” + “for” + “GeeksGeeks” ==> GeeksforGeeksGeeks. so the character at 3rd position is ‘e’.
Approach: The simple solution would be to keep appending strings in the string str, until the index reaches k. Then, return the character to the index k.
Below is the implementation of the above approach.
C++
// C++ program to solve the above approach#include<bits/stdc++.h>using namespace std;
char KthCharacter(string s, string t, long k)
{ // initializing first and second variable
// as to store how many string 's'
// and string 't' will be appended
long f = 1;
long ss = 2;
string tmp = "";
// storing tmp length
int len = tmp.length();
// if length of string tmp is greater than k, then
// we have reached our destination string now we can
// return character at index k
while (len < k) {
long tf = f;
long ts = ss;
// appending s to tmp, f times
while (tf-- != 0) {
tmp += s;
}
// appending t to tmp, s times
while (ts-- != 0) {
tmp += t;
}
f += 2;
ss += 2;
len = tmp.length();
}
// extracting output character
char output = tmp[k - 1];
return output;
}// Driver codeint main()
{ string S1 = "a", S2 = "bc";
int k = 4;
char ans = KthCharacter(S1, S2, k);
cout<<ans;
}// This code is contributed by ipg2016107. |
Java
// Java program to solve the above approachimport java.io.*;
class GFG {
static char KthCharacter(String s, String t, long k)
{
// initializing first and second variable
// as to store how many string 's'
// and string 't' will be appended
long f = 1;
long ss = 2;
String tmp = "";
// storing tmp length
int len = tmp.length();
// if length of string tmp is greater than k, then
// we have reached our destination string now we can
// return character at index k
while (len < k) {
long tf = f;
long ts = ss;
// appending s to tmp, f times
while (tf-- != 0) {
tmp += s;
}
// appending t to tmp, s times
while (ts-- != 0) {
tmp += t;
}
f += 2;
ss += 2;
len = tmp.length();
}
// extracting output character
char output = tmp.charAt((int)k - 1);
return output;
}
// Driver code
public static void main(String[] args)
{
String S1 = "a", S2 = "bc";
int k = 4;
char ans = KthCharacter(S1, S2, k);
System.out.println(ans);
}
} |
Python3
# Python program to solve the above approachdef KthCharacter(s, t, k):
# initializing first and second variable
# as to store how many string 's'
# and string 't' will be appended
f = 1
ss = 2
tmp = ""
# storing tmp length
lenn = len(tmp)
# if length of string tmp is greater than k, then
# we have reached our destination string now we can
# return character at index k
while (lenn < k):
tf = f
ts = ss
# appending s to tmp, f times
while (tf != 0):
tf -=1
tmp += s
# appending t to tmp, s times
while (ts != 0) :
ts -=1
tmp += t
f += 2
ss += 2
lenn = len(tmp)
# extracting output character
output = tmp[k - 1]
return output
# Driver codeS1 = "a"
S2 = "bc"
k = 4
ans = KthCharacter(S1, S2, k)
print(ans)
# This code is contributed by shivanisinghss2110 |
C#
// C# program to solve the above approachusing System;
class GFG {
static char KthCharacter(string s, string t, long k)
{
// initializing first and second variable
// as to store how many string 's'
// and string 't' will be appended
long f = 1;
long ss = 2;
string tmp = "";
// storing tmp length
int len = tmp.Length;
// if length of string tmp is greater than k, then
// we have reached our destination string now we can
// return character at index k
while (len < k) {
long tf = f;
long ts = ss;
// appending s to tmp, f times
while (tf-- != 0) {
tmp += s;
}
// appending t to tmp, s times
while (ts-- != 0) {
tmp += t;
}
f += 2;
ss += 2;
len = tmp.Length;
}
// extracting output character
char output = tmp[(int)k - 1];
return output;
}
// Driver code
public static void Main(string[] args)
{
string S1 = "a", S2 = "bc";
int k = 4;
char ans = KthCharacter(S1, S2, k);
Console.Write(ans);
}
}// This code is contributed by ukasp. |
Javascript
<script>// JavaScript program to solve the above approachfunction KthCharacter(s, t, k) {
// initializing first and second variable
// as to store how many string 's'
// and string 't' will be appended
let f = 1;
let ss = 2;
let tmp = "";
// storing tmp length
let len = tmp.length;
// if length of string tmp is greater than k, then
// we have reached our destination string now we can
// return character at index k
while (len < k) {
let tf = f;
let ts = ss;
// appending s to tmp, f times
while (tf-- != 0) {
tmp += s;
}
// appending t to tmp, s times
while (ts-- != 0) {
tmp += t;
}
f += 2;
ss += 2;
len = tmp.length;
}
// extracting output character
let output = tmp[k - 1];
return output;
}// Driver codelet S1 = "a",
S2 = "bc";
let k = 4;let ans = KthCharacter(S1, S2, k);document.write(ans);</script> |
Output
b
Time Complexity: O(K)
Auxiliary Space: O(K)
Efficient Approach: Since K is a long type, and the size of a string can only get up to the max possible value of int, this code cannot be used and won’t pass in every case. Rather than appending the string s and t to temp, subtract the length of s and length of t from K, so that the range is always within int type. Follow the steps below to solve the problem:
- Initialize two variables n and m as the lengths of the respective strings s and t.
- Initialize two variables, first and second as 1 and 2 to keep the count of time strings s and t have to be appended.
- Initialize the variable output to store the character at kth position in the new string.
-
Iterate in a while loop till k is greater than 0:
- If k is greater than first*n, then, subtract that from k and increase the value of first by 2.
- If k is greater than second*m, then, subtract that from k and increase the value of second by 2.
- Else(k is less than second*m), then the character from the second string t will be the answer as the string t will be appended second times and k will be in this range of positions.
- Initialize the variable ind as the modulus of k%m and the character at ind position in the string t will be the answer. Store the character in the variable output.
- Else(k is less than first*n), then the character from the second string s will be the answer as the string s will be appended first times and k will be in this range of positions.
- Initialize the variable ind as the modulus of k%n and the character at ind position in the string s will be the answer. Store the character in the variable output.
- After performing the above steps, return the value of output as the answer.
Below is the implementation of the above approach.
C++
// C++ program to solve the above approach#include<bits/stdc++.h>using namespace std;
char KthCharacter(string s, string t, long k)
{ // storing length
long n = s.length();
long m = t.length();
// variables for temporary strings of s and t
long first = 1;
long second = 2;
// final output variable
char output = '?';
while (k > 0) {
// if k is greater than even after adding string
// 's' 'first' times (let's call it x) we'll
// subtract x from k
if (k > first * n) {
long x = first * n;
k = k - x;
// incrementing first by 2, as said in
// problem statement
first = first + 2;
// if k is greater than even after adding
// string 't' 'second' times (let's call it
// y) we'll subtract y from k
if (k > second * m) {
long y = second * m;
k = k - y;
// incrementing second by two as said in
// problem statement
second = second + 2;
}
// if k is smaller than y, then the
// character
// will be at position k%m.
else {
// returning character at k index
long ind = k % m;
ind--;
if (ind < 0)
ind = m - 1;
output = t[(int)ind];
break;
}
}
// if k is smaller than x, then the character
// is at position k%n
else {
// returning character at k index
long ind = k % n;
ind--;
if (ind < 0)
ind = n - 1;
output = s[(int)ind];
break;
}
}
return output;
}// Driver codeint main()
{ string S1 = "a", S2 = "bc";
int k = 4;
char ans = KthCharacter(S1, S2, k);
cout<<(ans);
return 0;
}// This code is contributed by umadevi9616 |
Java
// Java program to solve the above approachimport java.io.*;
class GFG {
static char KthCharacter(String s, String t, long k)
{
// storing length
long n = s.length();
long m = t.length();
// variables for temporary strings of s and t
long first = 1;
long second = 2;
// final output variable
char output = '?';
while (k > 0) {
// if k is greater than even after adding string
// 's' 'first' times (let's call it x) we'll
// subtract x from k
if (k > first * n) {
long x = first * n;
k = k - x;
// incrementing first by 2, as said in
// problem statement
first = first + 2;
// if k is greater than even after adding
// string 't' 'second' times (let's call it
// y) we'll subtract y from k
if (k > second * m) {
long y = second * m;
k = k - y;
// incrementing second by two as said in
// problem statement
second = second + 2;
}
// if k is smaller than y, then the
// character
// will be at position k%m.
else {
// returning character at k index
long ind = k % m;
ind--;
if (ind < 0)
ind = m - 1;
output = t.charAt((int)ind);
break;
}
}
// if k is smaller than x, then the character
// is at position k%n
else {
// returning character at k index
long ind = k % n;
ind--;
if (ind < 0)
ind = n - 1;
output = s.charAt((int)ind);
break;
}
}
return output;
}
// Driver code
public static void main(String[] args)
{
String S1 = "a", S2 = "bc";
int k = 4;
char ans = KthCharacter(S1, S2, k);
System.out.println(ans);
}
} |
Python3
# Python program to solve the above approachdef KthCharacter(s, t, k):
# storing length
n = len(s)
m = len(t)
# variables for temporary strings of s and t
first = 1
second = 2
# final output variable
output = '?'
while (k > 0):
# if k is greater than even after adding string
# 's' 'first' times (let's call it x) we'll
# subtract x from k
if (k > first * n):
x = first * n
k = k - x
# incrementing first by 2, as said in
# problem statement
first = first + 2
# if k is greater than even after adding
# string 't' 'second' times (let's call it
# y) we'll subtract y from k
if (k > second * m):
y = second * m
k = k - y
# incrementing second by two as said in
# problem statement
second = second + 2
# if k is smaller than y, then the
# character
# will be at position k%m.
else:
# returning character at k index
ind = k % m
ind-=1
if (ind < 0):
ind = m - 1
output = t[ind]
break
# if k is smaller than x, then the character
# is at position k%n
else:
# returning character at k index
ind = k % n
ind-=1
if (ind < 0):
ind = n - 1
output = s[ind]
break
return output
# Driver codeS1 = "a"
S2 = "bc"
k = 4
ans = KthCharacter(S1, S2, k)
print(ans)
# This code is contributed by shivanisinghss2110 |
C#
// C# program to solve the above approachusing System;
public class GFG {
static char Kthchar(String s, String t, long k)
{
// storing length
long n = s.Length;
long m = t.Length;
// variables for temporary strings of s and t
long first = 1;
long second = 2;
// readonly output variable
char output = '?';
while (k > 0) {
// if k is greater than even after adding string
// 's' 'first' times (let's call it x) we'll
// subtract x from k
if (k > first * n) {
long x = first * n;
k = k - x;
// incrementing first by 2, as said in
// problem statement
first = first + 2;
// if k is greater than even after adding
// string 't' 'second' times (let's call it
// y) we'll subtract y from k
if (k > second * m) {
long y = second * m;
k = k - y;
// incrementing second by two as said in
// problem statement
second = second + 2;
}
// if k is smaller than y, then the
// character
// will be at position k%m.
else {
// returning character at k index
long ind = k % m;
ind--;
if (ind < 0)
ind = m - 1;
output = t[(int)(ind)];
break;
}
}
// if k is smaller than x, then the character
// is at position k%n
else {
// returning character at k index
long ind = k % n;
ind--;
if (ind < 0)
ind = n - 1;
output = s[(int)(ind)];
break;
}
}
return output;
}
// Driver code
public static void Main(String[] args)
{
String S1 = "a", S2 = "bc";
int k = 4;
char ans = Kthchar(S1, S2, k);
Console.WriteLine(ans);
}
}// This code is contributed by Amit Katiyar |
Javascript
<script>// javascript program to solve the above approach function KthCharacter(s, t , k)
{
// storing length
var n = s.length;
var m = t.length;
// variables for temporary strings of s and t
var first = 1;
var second = 2;
// final output variable
var output = '?';
while (k > 0) {
// if k is greater than even after adding string
// 's' 'first' times (let's call it x) we'll
// subtract x from k
if (k > first * n) {
var x = first * n;
k = k - x;
// incrementing first by 2, as said in
// problem statement
first = first + 2;
// if k is greater than even after adding
// string 't' 'second' times (let's call it
// y) we'll subtract y from k
if (k > second * m) {
var y = second * m;
k = k - y;
// incrementing second by two as said in
// problem statement
second = second + 2;
}
// if k is smaller than y, then the
// character
// will be at position k%m.
else {
// returning character at k index
var ind = k % m;
ind--;
if (ind < 0)
ind = m - 1;
output = t.charAt(parseInt(ind));
break;
}
}
// if k is smaller than x, then the character
// is at position k%n
else {
// returning character at k index
var ind = k % n;
ind--;
if (ind < 0)
ind = n - 1;
output = s.charAt(parseInt(ind));
break;
}
}
return output;
}
// Driver code
var S1 = "a", S2 = "bc";
var k = 4;
var ans = KthCharacter(S1, S2, k);
document.write(ans);
// This code contributed by Princi Singh </script> |
Output
b
Time Complexity: O(K)
Auxiliary Space: O(1)
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