Minimum length substring with each letter occurring both in uppercase and lowercase
Last Updated :
12 Aug, 2022
Find the minimum length substring in the given string str such that, in the substring, each alphabet appears at least once in lower case and at least once in upper case.
Examples:
Input: S = “AaBbCc”
Output: Aa
Explanation: Possible substrings that has each alphabet in lowercase and uppercase are:
Aa
Bb
Cc
AaBb
BbCc
AaBbCc
Among these, the minimum length substrings are Aa, Bb and Cc. Hence any of them can be a possible answer.
Input: S = "Geeks"
Output: -1
Explanation: No such substring present.
Naive Approach: The most straightforward approach is to generate all possible substrings of the given string and check if there exists any substring satisfying the given conditions. Print the smallest of all such substrings.
Time Complexity: O(N3)
Auxiliary Space: O(N)
Efficient Approach: To optimize the above approach, the idea is to use the concept of Sliding Window. Follow the steps below to solve the problem:
- Traverse the given string and store the characters whose only lowercase or uppercase form are present in the input string in a Map mp.
- Initialize two arrays to keep track of the lowercase and uppercase characters obtained so far.
- Now, traverse the string maintaining two pointers i and st (initialized with 0), where st will point to the start of the current substring and i will point to the current character.
- If the current character is in mp, neglect all the characters obtained so far and start from the next character and adjust the arrays accordingly.
- If the current character is not in mp, remove the extra characters from the beginning of the substring with the help of st pointer, such that the frequency of any character does not convert to 0 and adjust the arrays accordingly.
- Now, check whether the substring {S[st], ....., S[i]} is balanced or not. If balanced and i - st + 1 is smaller than the length of balanced substring obtained so far. Update the length and also store the start and end indices of the substring, i.e. st and i respectively.
- Repeat the steps till the end of the string.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if the current
// string is balanced or not
bool balanced(int small[], int caps[])
{
// For every character, check if
// there exists uppercase as well
// as lowercase characters
for (int i = 0; i < 26; i++) {
if (small[i] != 0 && (caps[i] == 0))
return 0;
else if ((small[i] == 0) && (caps[i] != 0))
return 0;
}
return 1;
}
// Function to find smallest length substring
// in the given string which is balanced
void smallestBalancedSubstring(string s)
{
// Store frequency of
// lowercase characters
int small[26];
// Stores frequency of
// uppercase characters
int caps[26];
memset(small, 0, sizeof(small));
memset(caps, 0, sizeof(caps));
// Count frequency of characters
for (int i = 0; i < s.length(); i++) {
if (s[i] >= 65 && s[i] <= 90)
caps[s[i] - 'A']++;
else
small[s[i] - 'a']++;
}
// Mark those characters which
// are not present in both
// lowercase and uppercase
unordered_map<char, int> mp;
for (int i = 0; i < 26; i++) {
if (small[i] && !caps[i])
mp[char(i + 'a')] = 1;
else if (caps[i] && !small[i])
mp[char(i + 'A')] = 1;
}
// Initialize the frequencies
// back to 0
memset(small, 0, sizeof(small));
memset(caps, 0, sizeof(caps));
// Marks the start and
// end of current substring
int i = 0, st = 0;
// Marks the start and end
// of required substring
int start = -1, end = -1;
// Stores the length of
// smallest balanced substring
int minm = INT_MAX;
while (i < s.length()) {
if (mp[s[i]]) {
// Remove all characters
// obtained so far
while (st < i) {
if (s[st] >= 65 && s[st] <= 90)
caps[s[st] - 'A']--;
else
small[s[st] - 'a']--;
st++;
}
i += 1;
st = i;
}
else {
if (s[i] >= 65 && s[i] <= 90)
caps[s[i] - 'A']++;
else
small[s[i] - 'a']++;
// Remove extra characters from
// front of the current substring
while (1) {
if (s[st] >= 65 && s[st] <= 90
&& caps[s[st] - 'A'] > 1) {
caps[s[st] - 'A']--;
st++;
}
else if (s[st] >= 97 && s[st] <= 122
&& small[s[st] - 'a'] > 1) {
small[s[st] - 'a']--;
st++;
}
else
break;
}
// If substring (st, i) is balanced
if (balanced(small, caps)) {
if (minm > (i - st + 1)) {
minm = i - st + 1;
start = st;
end = i;
}
}
i += 1;
}
}
// No balanced substring
if (start == -1 || end == -1)
cout << -1 << endl;
// Store answer string
else {
string ans = "";
for (int i = start; i <= end; i++)
ans += s[i];
cout << ans << endl;
}
}
// Driver Code
int main()
{
// Given string
string s = "azABaabba";
smallestBalancedSubstring(s);
return 0;
}
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG{
// Function to check if the current
// string is balanced or not
static boolean balanced(int small[],
int caps[])
{
// For every character, check if
// there exists uppercase as well
// as lowercase characters
for(int i = 0; i < 26; i++)
{
if (small[i] != 0 && (caps[i] == 0))
return false;
else if ((small[i] == 0) && (caps[i] != 0))
return false;
}
return true;
}
// Function to find smallest length substring
// in the given string which is balanced
static void smallestBalancedSubstring(String s)
{
// Store frequency of
// lowercase characters
int[] small = new int[26];
// Stores frequency of
// uppercase characters
int[] caps = new int[26];
Arrays.fill(small, 0);
Arrays.fill(caps, 0);
// Count frequency of characters
for(int i = 0; i < s.length(); i++)
{
if (s.charAt(i) >= 65 && s.charAt(i) <= 90)
caps[s.charAt(i) - 'A']++;
else
small[s.charAt(i) - 'a']++;
}
// Mark those characters which
// are not present in both
// lowercase and uppercase
Map<Character,
Integer> mp = new HashMap<Character,
Integer>();
for(int i = 0; i < 26; i++)
{
if (small[i] != 0 && caps[i] == 0)
mp.put((char)(i + 'a'), 1);
else if (caps[i] != 0 && small[i] == 0)
mp.put((char)(i + 'A'), 1);
// mp[char(i + 'A')] = 1;
}
// Initialize the frequencies
// back to 0
Arrays.fill(small, 0);
Arrays.fill(caps, 0);
// Marks the start and
// end of current substring
int i = 0, st = 0;
// Marks the start and end
// of required substring
int start = -1, end = -1;
// Stores the length of
// smallest balanced substring
int minm = Integer.MAX_VALUE;
while (i < s.length())
{
if (mp.get(s.charAt(i)) != null)
{
// Remove all characters
// obtained so far
while (st < i)
{
if (s.charAt(st) >= 65 &&
s.charAt(st) <= 90)
caps[s.charAt(st) - 'A']--;
else
small[s.charAt(st) - 'a']--;
st++;
}
i += 1;
st = i;
}
else
{
if (s.charAt(i) >= 65 && s.charAt(i) <= 90)
caps[s.charAt(i) - 'A']++;
else
small[s.charAt(i) - 'a']++;
// Remove extra characters from
// front of the current substring
while (true)
{
if (s.charAt(st) >= 65 &&
s.charAt(st) <= 90 &&
caps[s.charAt(st) - 'A'] > 1)
{
caps[s.charAt(st) - 'A']--;
st++;
}
else if (s.charAt(st) >= 97 &&
s.charAt(st) <= 122 &&
small[s.charAt(st) - 'a'] > 1)
{
small[s.charAt(st) - 'a']--;
st++;
}
else
break;
}
// If substring (st, i) is balanced
if (balanced(small, caps))
{
if (minm > (i - st + 1))
{
minm = i - st + 1;
start = st;
end = i;
}
}
i += 1;
}
}
// No balanced substring
if (start == -1 || end == -1)
System.out.println(-1);
// Store answer string
else
{
String ans = "";
for(int j = start; j <= end; j++)
ans += s.charAt(j);
System.out.println(ans);
}
}
// Driver Code
public static void main(String[] args)
{
// Given string
String s = "azABaabba";
smallestBalancedSubstring(s);
}
}
// This code is contributed by Dharanendra L V
# python 3 program for the above approach
import sys
# Function to check if the current
# string is balanced or not
def balanced(small, caps):
# For every character, check if
# there exists uppercase as well
# as lowercase characters
for i in range(26):
if (small[i] != 0 and (caps[i] == 0)):
return 0
elif((small[i] == 0) and (caps[i] != 0)):
return 0
return 1
# Function to find smallest length substring
# in the given string which is balanced
def smallestBalancedSubstring(s):
# Store frequency of
# lowercase characters
small = [0 for i in range(26)]
# Stores frequency of
# uppercase characters
caps = [0 for i in range(26)]
# Count frequency of characters
for i in range(len(s)):
if (ord(s[i]) >= 65 and ord(s[i]) <= 90):
caps[ord(s[i]) - 65] += 1
else:
small[ord(s[i]) - 97] += 1
# Mark those characters which
# are not present in both
# lowercase and uppercase
mp = {}
for i in range(26):
if (small[i] and caps[i]==0):
mp[chr(i + 97)] = 1
elif (caps[i] and small[i]==0):
mp[chr(i + 65)] = 1
# Initialize the frequencies
# back to 0
for i in range(len(small)):
small[i] = 0
caps[i] = 0
# Marks the start and
# end of current substring
i = 0
st = 0
# Marks the start and end
# of required substring
start = -1
end = -1
# Stores the length of
# smallest balanced substring
minm = sys.maxsize
while (i < len(s)):
if(s[i] in mp):
# Remove all characters
# obtained so far
while (st < i):
if (ord(s[st]) >= 65 and ord(s[st]) <= 90):
caps[ord(s[st]) - 65] -= 1
else:
small[ord(s[st]) - 97] -= 1
st += 1
i += 1
st = i
else:
if (ord(s[i]) >= 65 and ord(s[i]) <= 90):
caps[ord(s[i]) - 65] += 1
else:
small[ord(s[i] )- 97] += 1
# Remove extra characters from
# front of the current substring
while (1):
if (ord(s[st]) >= 65 and ord(s[st])<= 90 and caps[ord(s[st])- 65] > 1):
caps[ord(s[st]) - 65] -= 1
st += 1
elif (ord(s[st]) >= 97 and ord(s[st]) <= 122 and small[ord(s[st]) - 97] > 1):
small[ord(s[st]) - 97] -= 1
st += 1
else:
break
# If substring (st, i) is balanced
if (balanced(small, caps)):
if (minm > (i - st + 1)):
minm = i - st + 1
start = st
end = i
i += 1
# No balanced substring
if (start == -1 or end == -1):
print(-1)
# Store answer string
else:
ans = ""
for i in range(start,end+1,1):
ans += s[i]
print(ans)
# Driver Code
if __name__ == '__main__':
# Given string
s = "azABaabba"
smallestBalancedSubstring(s)
# This code is contributed by bgangwar59.
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
public const int MaxValue = 2147483647;
// Function to check if the current
// string is balanced or not
static bool balanced(int []small,
int []caps)
{
// For every character, check if
// there exists uppercase as well
// as lowercase characters
for(int i = 0; i < 26; i++)
{
if (small[i] != 0 && (caps[i] == 0))
return false;
else if ((small[i] == 0) && (caps[i] != 0))
return false;
}
return true;
}
// Function to find smallest length substring
// in the given string which is balanced
static void smallestBalancedSubstring(string s)
{
// Store frequency of
// lowercase characters
int[] small = new int[26];
int i;
// Stores frequency of
// uppercase characters
int[] caps = new int[26];
Array.Clear(small, 0, small.Length);
Array.Clear(caps, 0, caps.Length);
// Count frequency of characters
for(i = 0; i < s.Length; i++)
{
if (s[i] >= 65 && s[i] <= 90)
caps[(int)s[i] - 65]++;
else
small[(int)s[i]- 97]++;
}
// Mark those characters which
// are not present in both
// lowercase and uppercase
Dictionary<char,int> mp = new Dictionary<char,int>();
for(i = 0; i < 26; i++)
{
if (small[i] != 0 && caps[i] == 0){
mp[(char)(i+97)] = 1;
}
else if (caps[i] != 0 && small[i] == 0)
mp[(char)(i+65)] = 1;
// mp[char(i + 'A')] = 1;
}
// Initialize the frequencies
// back to 0
Array.Clear(small, 0, small.Length);
Array.Clear(caps, 0, caps.Length);
// Marks the start and
// end of current substring
i = 0;
int st = 0;
// Marks the start and end
// of required substring
int start = -1, end = -1;
// Stores the length of
// smallest balanced substring
int minm = MaxValue;
while (i < s.Length)
{
if (mp.ContainsKey(s[i]))
{
// Remove all characters
// obtained so far
while (st < i)
{
if ((int)s[st] >= 65 &&
(int)s[st] <= 90)
caps[(int)s[st] - 65]--;
else
small[(int)s[st] - 97]--;
st++;
}
i += 1;
st = i;
}
else
{
if ((int)s[i] >= 65 && (int)s[i] <= 90)
caps[(int)s[i] - 65]++;
else
small[(int)s[i] - 97]++;
// Remove extra characters from
// front of the current substring
while (true)
{
if ((int)s[st] >= 65 &&
(int)s[st] <= 90 &&
caps[(int)s[st] - 65] > 1)
{
caps[(int)s[st] - 65]--;
st++;
}
else if ((int)s[st] >= 97 &&
(int)s[st] <= 122 &&
small[(int)s[st] - 97] > 1)
{
small[(int)s[st] - 97]--;
st++;
}
else
break;
}
// If substring (st, i) is balanced
if (balanced(small, caps))
{
if (minm > (i - st + 1))
{
minm = i - st + 1;
start = st;
end = i;
}
}
i += 1;
}
}
// No balanced substring
if (start == -1 || end == -1)
Console.WriteLine(-1);
// Store answer string
else
{
string ans = "";
for(int j = start; j <= end; j++)
ans += s[j];
Console.WriteLine(ans);
}
}
// Driver Code
public static void Main()
{
// Given string
string s = "azABaabba";
smallestBalancedSubstring(s);
}
}
// This code is contributed by SURENDRA_GANGWAR.
<script>
// Javascript program for the above approach
let MaxValue = 2147483647;
// Function to check if the current
// string is balanced or not
function balanced(small, caps)
{
// For every character, check if
// there exists uppercase as well
// as lowercase characters
for(let i = 0; i < 26; i++)
{
if (small[i] != 0 && (caps[i] == 0))
return false;
else if ((small[i] == 0) && (caps[i] != 0))
return false;
}
return true;
}
// Function to find smallest length substring
// in the given string which is balanced
function smallestBalancedSubstring(s)
{
// Store frequency of
// lowercase characters
let small = new Array(26);
let i;
// Stores frequency of
// uppercase characters
let caps = new Array(26);
small.fill(0);
caps.fill(0);
// Count frequency of characters
for(i = 0; i < s.length; i++)
{
if (s[i].charCodeAt() >= 65 && s[i].charCodeAt() <= 90)
caps[s[i].charCodeAt() - 65]++;
else
small[s[i].charCodeAt()- 97]++;
}
// Mark those characters which
// are not present in both
// lowercase and uppercase
let mp = new Map();
for(i = 0; i < 26; i++)
{
if (small[i] != 0 && caps[i] == 0){
mp.set(String.fromCharCode(i+97), 1);
}
else if (caps[i] != 0 && small[i] == 0)
mp.set(String.fromCharCode(i+65), 1);
// mp[char(i + 'A')] = 1;
}
// Initialize the frequencies
// back to 0
small.fill(0);
caps.fill(0);
// Marks the start and
// end of current substring
i = 0;
let st = 0;
// Marks the start and end
// of required substring
let start = -1, end = -1;
// Stores the length of
// smallest balanced substring
let minm = MaxValue;
while (i < s.length)
{
if (mp.has(s[i]))
{
// Remove all characters
// obtained so far
while (st < i)
{
if (s[st].charCodeAt() >= 65 &&
s[st].charCodeAt() <= 90)
caps[s[st].charCodeAt() - 65]--;
else
small[s[st].charCodeAt() - 97]--;
st++;
}
i += 1;
st = i;
}
else
{
if (s[i].charCodeAt() >= 65 && s[i].charCodeAt() <= 90)
caps[s[i].charCodeAt() - 65]++;
else
small[s[i].charCodeAt() - 97]++;
// Remove extra characters from
// front of the current substring
while (true)
{
if (s[st].charCodeAt() >= 65 &&
s[st].charCodeAt() <= 90 &&
caps[s[st].charCodeAt() - 65] > 1)
{
caps[s[st].charCodeAt() - 65]--;
st++;
}
else if (s[st].charCodeAt() >= 97 &&
s[st].charCodeAt() <= 122 &&
small[s[st].charCodeAt() - 97] > 1)
{
small[s[st].charCodeAt() - 97]--;
st++;
}
else
break;
}
// If substring (st, i) is balanced
if (balanced(small, caps))
{
if (minm > (i - st + 1))
{
minm = i - st + 1;
start = st;
end = i;
}
}
i += 1;
}
}
// No balanced substring
if (start == -1 || end == -1)
document.write(-1 + "</br>");
// Store answer string
else
{
let ans = "";
for(let j = start; j <= end; j++)
ans += s[j];
document.write(ans + "</br>");
}
}
// Given string
let s = "azABaabba";
smallestBalancedSubstring(s);
// This code is contributed by decode2207.
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
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