Alternate vowel and consonant string
Last Updated :
15 Mar, 2023
Given a string, rearrange the characters of the given string such that the vowels and consonants occupy the alternate position. If the string can not be rearranged in the desired way, print "no such string". The order of vowels with respect to each other and the order of consonants with respect to each other should be maintained.
If more than one required string can be formed, print the lexicographically smaller.
Examples:
Input : geeks
Output : gekes
Input : onse
Output : nose
There are two possible outcomes
"nose" and "ones". Since "nose"
is lexicographically smaller, we
print it.
- Count number of vowels and consonants in given string.
- If difference between counts is more than one, return "Not Possible".
- If there are more vowels than consonants, print the first vowel first and recur for the remaining string.
- If there are more consonants than vowels, print the first consonant first and recur for the remaining string.
- If counts are same, compare first vowel with first consonant and print the smaller one first.
Implementation:
C++
// C++ implementation of alternate vowel and
// consonant string
#include <bits/stdc++.h>
using namespace std;
// 'ch' is vowel or not
bool isVowel(char ch)
{
if (ch == 'a' || ch == 'e' || ch == 'i' ||
ch == 'o' || ch =='u')
return true;
return false;
}
// create alternate vowel and consonant string
// str1[0...l1-1] and str2[start...l2-1]
string createAltStr(string str1, string str2,
int start, int l)
{
string finalStr = "";
// first adding character of vowel/consonant
// then adding character of consonant/vowel
for (int i=0, j=start; j<l; i++, j++)
finalStr = (finalStr + str1.at(i)) + str2.at(j);
return finalStr;
}
// function to find the required
// alternate vowel and consonant string
string findAltStr(string str)
{
int nv = 0, nc = 0;
string vstr = "", cstr = "";
int l = str.size();
for (int i=0; i<l; i++)
{
char ch = str.at(i);
// count vowels and update vowel string
if (isVowel(ch))
{
nv++;
vstr = vstr + ch;
}
// count consonants and update consonant
// string
else
{
nc++;
cstr = cstr + ch;
}
}
// no such string can be formed
if (abs(nv-nc) >= 2)
return "no such string";
// remove first character of vowel string
// then create alternate string with
// cstr[0...nc-1] and vstr[1...nv-1]
if (nv > nc)
return (vstr.at(0) + createAltStr(cstr, vstr, 1, nv));
// remove first character of consonant string
// then create alternate string with
// vstr[0...nv-1] and cstr[1...nc-1]
if (nc > nv)
return (cstr.at(0) + createAltStr(vstr, cstr, 1, nc));
// if both vowel and consonant
// strings are of equal length
// start creating string with consonant
if (cstr.at(0) < vstr.at(0))
return createAltStr(cstr, vstr, 0, nv);
// start creating string with vowel
return createAltStr(vstr, cstr, 0, nc);
}
// Driver program to test above
int main()
{
string str = "geeks";
cout << findAltStr(str);
return 0;
}
Java
// Java implementation of alternate vowel and
// consonant string
import java.util.*;
class GFG
{
// 'ch' is vowel or not
static boolean isVowel(char ch)
{
if (ch == 'a' || ch == 'e' || ch == 'i' ||
ch == 'o' || ch =='u')
return true;
return false;
}
// create alternate vowel and consonant string
// str1[0...l1-1] and str2[start...l2-1]
static String createAltStr(String str1, String str2,
int start, int l)
{
String finalStr = "";
// first adding character of vowel/consonant
// then adding character of consonant/vowel
for (int i = 0, j = start; j < l; i++, j++)
finalStr = (finalStr + str1.charAt(i)) +
str2.charAt(j);
return finalStr;
}
// function to find the required
// alternate vowel and consonant string
static String findAltStr(String str)
{
int nv = 0, nc = 0;
String vstr = "", cstr = "";
int l = str.length();
for (int i = 0; i < l; i++)
{
char ch = str.charAt(i);
// count vowels and update vowel string
if (isVowel(ch))
{
nv++;
vstr = vstr + ch;
}
// count consonants and update consonant
// string
else
{
nc++;
cstr = cstr + ch;
}
}
// no such string can be formed
if (Math.abs(nv - nc) >= 2)
return "no such string";
// remove first character of vowel string
// then create alternate string with
// cstr[0...nc-1] and vstr[1...nv-1]
if (nv > nc)
return (vstr.charAt(0) + createAltStr(cstr, vstr, 1, nv));
// remove first character of consonant string
// then create alternate string with
// vstr[0...nv-1] and cstr[1...nc-1]
if (nc > nv)
return (cstr.charAt(0) + createAltStr(vstr, cstr, 1, nc));
// if both vowel and consonant
// strings are of equal length
// start creating string with consonant
if (cstr.charAt(0) < vstr.charAt(0))
return createAltStr(cstr, vstr, 0, nv);
// start creating string with vowel
return createAltStr(vstr, cstr, 0, nc);
}
// Driver code
public static void main(String args[])
{
String str = "geeks";
System.out.println(findAltStr(str));
}
}
// This code is contributed by
// Shashank_Sharma
Python 3
# Python implementation of alternate vowel
# and consonant string
# 'ch' is vowel or not
def isVowel(ch):
if(ch == 'a' or ch == 'e' or
ch == 'i' or ch == 'o' or
ch == 'u'):
return True
return False
# create alternate vowel and consonant string
# str1[0...l1-1] and str2[start...l2-1]
def createAltStr(str1, str2, start, l):
finalStr = ""
i = 0
# first adding character of vowel/consonant
# then adding character of consonant/vowel
for j in range(start, l):
finalStr = (finalStr + str1[i]) + str2[j]
i + 1
return finalStr
# function to find the required
# alternate vowel and consonant string
def findAltStr(str1):
nv = 0
nc = 0
vstr = ""
cstr = ""
l = len(str1)
for i in range(0, l):
# count vowels and update vowel string
if(isVowel(str1[i])):
nv += 1
vstr = vstr + str1[i]
# count consonants and update
# consonant string
else:
nc += 1
cstr = cstr + str1[i]
# no such string can be formed
if(abs(nv - nc) >= 2):
return "no such string"
# remove first character of vowel string
# then create alternate string with
# cstr[0...nc-1] and vstr[1...nv-1]
if(nv > nc):
return (vstr[0] + createAltStr(cstr, vstr, 1, nv))
# remove first character of consonant string
# then create alternate string with
# vstr[0...nv-1] and cstr[1...nc-1]
if(nc > nv):
return (cstr[0] + createAltStr(vstr, cstr, 1, nc))
# if both vowel and consonant
# strings are of equal length
# start creating string with consonant
if(cstr[0] < vstr[0]):
return createAltStr(cstr, vstr, 0, nv)
return createAltStr(vstr, cstr, 0, nc)
# Driver Code
if __name__ == "__main__":
str1 = "geeks"
print(findAltStr(str1))
# This code is contributed by Sairahul099
C#
// C# implementation of alternate vowel and
// consonant string
using System;
class GFG
{
// 'ch' is vowel or not
static Boolean isVowel(char ch)
{
if (ch == 'a' || ch == 'e' || ch == 'i' ||
ch == 'o' || ch =='u')
return true;
return false;
}
// create alternate vowel and consonant string
// str1[0...l1-1] and str2[start...l2-1]
static String createAltStr(String str1, String str2,
int start, int l)
{
String finalStr = "";
// first adding character of vowel/consonant
// then adding character of consonant/vowel
for (int i = 0, j = start; j < l; i++, j++)
finalStr = (finalStr + str1[i]) +
str2[j];
return finalStr;
}
// function to find the required
// alternate vowel and consonant string
static String findAltStr(String str)
{
int nv = 0, nc = 0;
String vstr = "", cstr = "";
int l = str.Length;
for (int i = 0; i < l; i++)
{
char ch = str[i];
// count vowels and update vowel string
if (isVowel(ch))
{
nv++;
vstr = vstr + ch;
}
// count consonants and update consonant
// string
else
{
nc++;
cstr = cstr + ch;
}
}
// no such string can be formed
if (Math.Abs(nv - nc) >= 2)
return "no such string";
// remove first character of vowel string
// then create alternate string with
// cstr[0...nc-1] and vstr[1...nv-1]
if (nv > nc)
return (vstr[0] + createAltStr(cstr, vstr, 1, nv));
// remove first character of consonant string
// then create alternate string with
// vstr[0...nv-1] and cstr[1...nc-1]
if (nc > nv)
return (cstr[0] + createAltStr(vstr, cstr, 1, nc));
// if both vowel and consonant
// strings are of equal length
// start creating string with consonant
if (cstr[0] < vstr[0])
return createAltStr(cstr, vstr, 0, nv);
// start creating string with vowel
return createAltStr(vstr, cstr, 0, nc);
}
// Driver code
public static void Main(String []args)
{
String str = "geeks";
Console.WriteLine(findAltStr(str));
}
}
// This code is contributed by Princi Singh
JavaScript
<script>
// JavaScript implementation of alternate vowel and
// consonant string
// 'ch' is vowel or not
function isVowel(ch)
{
if (ch == 'a' || ch == 'e' || ch == 'i' ||
ch == 'o' || ch =='u')
return true;
return false;
}
// create alternate vowel and consonant string
// str1[0...l1-1] and str2[start...l2-1]
function createAltStr(str1, str2,start,l)
{
let finalStr = "";
// first adding character of vowel/consonant
// then adding character of consonant/vowel
for (let i=0, j=start; j<l; i++, j++)
finalStr = (finalStr + str1[i] + str2[j]);
return finalStr;
}
// function to find the required
// alternate vowel and consonant string
function findAltStr(str)
{
let nv = 0, nc = 0;
let vstr = "", cstr = "";
let l = str.length;
for (let i=0; i<l; i++)
{
let ch = str[i];
// count vowels and update vowel string
if (isVowel(ch))
{
nv++;
vstr = vstr + ch;
}
// count consonants and update consonant
// string
else
{
nc++;
cstr = cstr + ch;
}
}
// no such string can be formed
if (Math.abs(nv-nc) >= 2)
return "no such string";
// remove first character of vowel string
// then create alternate string with
// cstr[0...nc-1] and vstr[1...nv-1]
if (nv > nc)
return (vstr[0] + createAltStr(cstr, vstr, 1, nv));
// remove first character of consonant string
// then create alternate string with
// vstr[0...nv-1] and cstr[1...nc-1]
if (nc > nv)
return (cstr[0] + createAltStr(vstr, cstr, 1, nc));
// if both vowel and consonant
// strings are of equal length
// start creating string with consonant
if (cstr.at(0) < vstr.at(0))
return createAltStr(cstr, vstr, 0, nv);
// start creating string with vowel
return createAltStr(vstr, cstr, 0, nc);
}
// Driver program to test above
let str = "geeks";
document.write(findAltStr(str));
// This code is contributed by Shinjan_Patra
</script>
Time Complexity: O(n), where 'n' the is length of the string
Auxiliary Space: O(n), where 'n' is the length of the string.
By Hashing:
The idea is to use the hash table to store the vowels and consonants occurrence then apply the simple brute force.
Steps to solve the problem:
1. declare the vector mp1 and mp2 to store the occurrence and variable v and c to store the count of vowel and consonants.
2. iterate through the string and increment the occurrence of vowel and consonants in the hash table.
3. if absolute difference of v and c is greater than 1, in this case string with alternate vowel and consonant is not possible so we will return "no such string".
4. declare variable it1 , it2 and i to iterate through the vector to find first vowel and consonants.
5. while it1<mp1.size() and mp1[it1] is equal to zero:
- we will increment the first iterator.
6. while it2<mp2.size() and mp2[it2] is equal to zero:
- we will increment the second iterator.
7. declare Boolean f to store c is greater than v or not to check either consonants will be at first place or the vowel.
8. if v is equal to c:
- f=it1>it2 (lexicographically smaller)
9. while it1 is smaller than mp1.size() and it2 is smaller than mp2.size() and i is smaller than n:
- if f is true s[i]=it2+'a' , --mp2[it2] and iterate through mp2 till mp2[it2] is equal to zero and increment it2, f=false.
- else s[i]=it1+'a' , --mp1[it1] and iterate through mp1 till mp1[it1] is equal to zero and increment it1, f=true.
- increment of i.
10. check for condition where only one vowel or consonant is left.
11. return the string.
Implementation of the approach:
C++
// C++ implementation of alternate vowel and
// consonant string
#include <bits/stdc++.h>
using namespace std;
string findAltStr(string s)
{
int n = s.size();
vector<int> mp1(26),
mp2(26); // to store vowels and consonants
int v = 0, c = 0;
for (char ch : s) {
if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o'
|| ch == 'u') { // if it's vowel
mp1[ch - 'a']++;
v++;
}
else { // consonant
mp2[ch - 'a']++;
c++;
}
}
if (abs(v - c) > 1)
return "no such string"; // if their diff is greater than one
// then string with alternate vowel and
// consonant cannot be made
int it1 = 0, it2 = 0, i = 0;
while (it1 < mp1.size() && mp1[it1] == 0)
it1++; // to find first vowel
while (it2 < mp2.size() && mp2[it2] == 0)
it2++; // to find first consonant
bool f
= c > v; // if number of consonant is greater then
// we will place consonant first else vowel
if (v == c) {
f = it1 > it2; // if both are equal then check which
// is lexiographically smaller
}
while ((it1 < mp1.size() && it2 < mp2.size())
&& i < n) {
if (f) {
s[i] = it2 + 'a';
--mp2[it2];
while (it2 < mp2.size() && mp2[it2] == 0)
it2++;
f = false; // this will trigger to place vowel
// next
}
else {
s[i] = it1 + 'a';
--mp1[it1];
while (it1 < mp1.size() && mp1[it1] == 0)
it1++;
f = true; // this will trigger to place
// consonant next
}
++i;
}
if (it1 != mp1.size())
s[i] = it1 + 'a'; // if one vowel left
else if (it2 != mp2.size())
s[i] = it2 + 'a'; // if one consonant left
return s;
}
// Driver program to test above
int main()
{
string str = "geeks";
cout << findAltStr(str);
return 0;
}
// This code is contributed by Prateek Kumar Singh
Java
// Java implementation of alternate vowel and
// consonant string
import java.util.*;
public class GFG {
static String findAltStr(String str)
{
char[] s = str.toCharArray();
int n = s.length;
int[] mp1 = new int[26];
int[] mp2
= new int[26]; // to store vowels and consonants
int v = 0, c = 0;
for (char ch : s) {
if (ch == 'a' || ch == 'e' || ch == 'i'
|| ch == 'o'
|| ch == 'u') { // if it's vowel
mp1[ch - 'a']++;
v++;
}
else { // consonant
mp2[ch - 'a']++;
c++;
}
}
if (Math.abs(v - c) > 1)
return "no such string"; // if their diff is
// greater than one
// then string with
// alternate vowel and
// consonant cannot be
// made
int it1 = 0, it2 = 0, i = 0;
while (it1 < mp1.length && mp1[it1] == 0)
it1++; // to find first vowel
while (it2 < mp2.length && mp2[it2] == 0)
it2++; // to find first consonant
boolean f = c > v; // if number of consonant is
// greater then we will place
// consonant first else vowel
if (v == c) {
f = it1
> it2; // if both are equal then check which
// is lexiographically smaller
}
while ((it1 < mp1.length && it2 < mp2.length)
&& i < n) {
if (f) {
s[i] = (char)(it2 + 'a');
--mp2[it2];
while (it2 < mp2.length && mp2[it2] == 0)
it2++;
f = false; // this will trigger to place
// vowel next
}
else {
s[i] = (char)(it1 + 'a');
--mp1[it1];
while (it1 < mp1.length && mp1[it1] == 0)
it1++;
f = true; // this will trigger to place
// consonant next
}
++i;
}
if (it1 != mp1.length)
s[i] = (char)(it1 + 'a'); // if one vowel left
else if (it2 != mp2.length)
s[i] = (char)(it2
+ 'a'); // if one consonant left
return String.valueOf(s);
}
// Driver program to test above
public static void main(String[] args)
{
String str = "geeks";
System.out.println(findAltStr(str));
}
}
// This code is contributed by Karandeep1234
C#
// C# implementation of alternate vowel and
// consonant string
using System;
class GFG {
static string FindAltStr(string str) {
char[] s = str.ToCharArray();
int n = s.Length;
int[] mp1 = new int[26];
int[] mp2 = new int[26]; // to store vowels and consonants
int v = 0, c = 0;
foreach (char ch in s) {
if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u') { // if it's vowel
mp1[ch - 'a']++;
v++;
}
else { // consonant
mp2[ch - 'a']++;
c++;
}
}
if (Math.Abs(v - c) > 1)
return "no such string"; // if their diff is
// greater than one
// then string with
// alternate vowel and
// consonant cannot be
// made
int it1 = 0, it2 = 0, i = 0;
while (it1 < mp1.Length && mp1[it1] == 0)
it1++; // to find first vowel
while (it2 < mp2.Length && mp2[it2] == 0)
it2++; // to find first consonant
bool f = c > v; // if number of consonant is
// greater then we will place
// consonant first else vowel
if (v == c) {
f = it1 > it2; // if both are equal then check which
// is lexiographically smaller
}
while ((it1 < mp1.Length && it2 < mp2.Length) && i < n) {
if (f) {
s[i] = (char)(it2 + 'a');
--mp2[it2];
while (it2 < mp2.Length && mp2[it2] == 0)
it2++;
f = false; // this will trigger to place
// vowel next
}
else {
s[i] = (char)(it1 + 'a');
--mp1[it1];
while (it1 < mp1.Length && mp1[it1] == 0)
it1++;
f = true; // this will trigger to place
// consonant next
}
++i;
}
if (it1 != mp1.Length)
s[i] = (char)(it1 + 'a'); // if one vowel left
else if (it2 != mp2.Length)
s[i] = (char)(it2 + 'a'); // if one consonant left
return new string(s);
}
// Driver program to test above
public static void Main(string[] args) {
string str = "geeks";
Console.WriteLine(FindAltStr(str));
}
}
// This code is contributed by Pushpesh Raj.
Python3
def findAltStr(s):
n = len(s)
mp1 = [0]*26
mp2 = [0]*26 # to store vowels and consonants
v = 0
c = 0
for ch in s:
if ch in ['a', 'e', 'i', 'o', 'u']: # if it's vowel
mp1[ord(ch) - ord('a')] += 1
v += 1
else: # consonant
mp2[ord(ch) - ord('a')] += 1
c += 1
if abs(v - c) > 1:
return "no such string" # if their diff is greater than one
# then string with alternate vowel and
# consonant cannot be made
it1 = 0
it2 = 0
i = 0
while it1 < len(mp1) and mp1[it1] == 0:
it1 += 1 # to find first vowel
while it2 < len(mp2) and mp2[it2] == 0:
it2 += 1 # to find first consonant
f = c > v # if number of consonant is greater then
# we will place consonant first else vowel
if v == c:
f = it1 > it2 # if both are equal then check which
# is lexiographically smaller
new_str = ['']*n
while it1 < len(mp1) and it2 < len(mp2) and i < n:
if f:
new_str[i] = chr(it2 + ord('a'))
mp2[it2] -= 1
while it2 < len(mp2) and mp2[it2] == 0:
it2 += 1
f = False # this will trigger to place vowel
# next
else:
new_str[i] = chr(it1 + ord('a'))
mp1[it1] -= 1
while it1 < len(mp1) and mp1[it1] == 0:
it1 += 1
f = True # this will trigger to place
# consonant next
i += 1
if it1 != len(mp1):
new_str[i] = chr(it1 + ord('a')) # if one vowel left
elif it2 != len(mp2):
new_str[i] = chr(it2 + ord('a')) # if one consonant left
return ''.join(new_str)
# Driver program to test above
str = "geeks"
print(findAltStr(str))
JavaScript
function findAltStr(s) {
const n = s.length;
const mp1 = new Array(26).fill(0);
const mp2 = new Array(26).fill(0); // to store vowels and consonants
let v = 0;
let c = 0;
for (const ch of s) {
if (['a', 'e', 'i', 'o', 'u'].includes(ch)) { // if it's vowel
mp1[ch.charCodeAt(0) - 'a'.charCodeAt(0)] += 1;
v += 1;
} else { // consonant
mp2[ch.charCodeAt(0) - 'a'.charCodeAt(0)] += 1;
c += 1;
}
}
if (Math.abs(v - c) > 1) {
return "no such string"; // if their diff is greater than one
// then string with alternate vowel and
// consonant cannot be made
}
let it1 = 0;
let it2 = 0;
let i = 0;
while (it1 < mp1.length && mp1[it1] == 0) {
it1 += 1; // to find first vowel
}
while (it2 < mp2.length && mp2[it2] == 0) {
it2 += 1; // to find first consonant
}
let f = c > v; // if number of consonant is greater then
// we will place consonant first else vowel
if (v === c) {
f = it1 > it2; // if both are equal then check which
// is lexiographically smaller
}
const new_str = new Array(n);
while (it1 < mp1.length && it2 < mp2.length && i < n) {
if (f) {
new_str[i] = String.fromCharCode(it2 + 'a'.charCodeAt(0));
mp2[it2] -= 1;
while (it2 < mp2.length && mp2[it2] === 0) {
it2 += 1;
}
f = false; // this will trigger to place vowel
// next
} else {
new_str[i] = String.fromCharCode(it1 + 'a'.charCodeAt(0));
mp1[it1] -= 1;
while (it1 < mp1.length && mp1[it1] === 0) {
it1 += 1;
}
f = true; // this will trigger to place
// consonant next
}
i += 1;
}
if (it1 !== mp1.length) {
new_str[i] = String.fromCharCode(it1 + 'a'.charCodeAt(0)); // if one vowel left
} else if (it2 !== mp2.length) {
new_str[i] = String.fromCharCode(it2 + 'a'.charCodeAt(0)); // if one consonant left
}
return new_str.join('');
}
// Driver program to test above
const str = "geeks";
console.log(findAltStr(str));
Time Complexity: O(n)
Auxiliary Space: O(n)
This approach is contributed by Prateek Kumar Singh (pkrsingh025).
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Basics & Prerequisites
Data Structures
Array Data StructureIn this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous
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String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut
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Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The
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Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Hereâs the comparison of Linked List vs Arrays Linked List:
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Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first
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Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems
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Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most
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Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of
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Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this
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Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
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Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
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Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
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Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
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Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
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Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
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Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
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Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
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Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
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GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
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