Given string **S** consists of lowercase and uppercase letters, the task is to find the number of substrings having an equal number of lowercase and uppercase letters.

**Examples:**

Input:S = “gEEk”Output:3Explanation:

The following are the substrings having an equal number of lowercase and uppercase letters:

- “gE”
- “gEEk”
- “Ek”
Therefore, the total count of substrings is 3.

Input:S = “WomensDAY”Output:4

**Naive Approach:** The simplest approach to solve the given problem is to generate all possible substring of the given string **S** and increment the count of a substring by **1** if that substring contains an equal number of lowercase and uppercase letters. After checking for all the substrings, print the value of the **count** as the result.

**Time Complexity:** O(N^{3})**Auxiliary Space:** O(1)

**Efficient Approach:** The given problem can be solved by considering each lowercase and uppercase letter as **1** and **-1** respectively, and then find the count of subarray having sum **0**. Follow the steps to solve the problem:

- Initialize a HashMap, say
**M**that stores the frequency of the sum of all the prefixes. - Initialize a variable, say,
**currentSum**as**0**and**res**as**0**that stores the sum of each prefix and count of the resultant substrings respectively. - Traverse the string and perform the following steps:
- If the current character is uppercase, then increment the value of
**currentSum**by**1**. Otherwise, decrease the value of**currentSum**by**-1**. - If the value of
**currentSum**is**0**, then increment the value of**res**by**1**. - If the value of
**currentSum**exists in the map**M**, then increment the value of**res**by**M[currentSum]**. - Increment the frequency of
**currentSum**in the HashMap**M**by**1**.

- If the current character is uppercase, then increment the value of
- After completing the above steps, print the value of
**res**as the result.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the count of` `// substrings having an equal number` `// of uppercase and lowercase characters` `int` `countSubstring(string& S, ` `int` `N)` `{` ` ` `// Stores the count of prefixes` ` ` `// having sum S considering uppercase` ` ` `// and lowercase characters as 1 and -1` ` ` `unordered_map<` `int` `, ` `int` `> prevSum;` ` ` `// Stores the count of substrings` ` ` `// having equal number of lowercase` ` ` `// and uppercase characters` ` ` `int` `res = 0;` ` ` `// Stores the sum obtained so far` ` ` `int` `currentSum = 0;` ` ` `for` `(` `int` `i = 0; i < N; i++) {` ` ` `// If the character is uppercase` ` ` `if` `(S[i] >= ` `'A'` `and S[i] <= ` `'Z'` `) {` ` ` `currentSum++;` ` ` `}` ` ` `// Otherwise` ` ` `else` ` ` `currentSum--;` ` ` `// If currsum is o` ` ` `if` `(currentSum == 0)` ` ` `res++;` ` ` `// If the current sum exists in` ` ` `// the HashMap prevSum` ` ` `if` `(prevSum.find(currentSum)` ` ` `!= prevSum.end()) {` ` ` `// Increment the resultant` ` ` `// count by 1` ` ` `res += (prevSum[currentSum]);` ` ` `}` ` ` `// Update the frequency of the` ` ` `// current sum by 1` ` ` `prevSum[currentSum]++;` ` ` `}` ` ` `// Return the resultant count of` ` ` `// the subarrays` ` ` `return` `res;` `}` `// Driver Code` `int` `main()` `{` ` ` `string S = ` `"gEEk"` `;` ` ` `cout << countSubstring(S, S.length());` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.util.HashMap;` `class` `GFG{` `// Function to find the count of` `// substrings having an equal number` `// of uppercase and lowercase characters` `static` `int` `countSubstring(String S, ` `int` `N)` `{` ` ` ` ` `// Stores the count of prefixes` ` ` `// having sum S considering uppercase` ` ` `// and lowercase characters as 1 and -1` ` ` `HashMap<Integer, Integer> prevSum = ` `new` `HashMap<>();` ` ` `// Stores the count of substrings` ` ` `// having equal number of lowercase` ` ` `// and uppercase characters` ` ` `int` `res = ` `0` `;` ` ` `// Stores the sum obtained so far` ` ` `int` `currentSum = ` `0` `;` ` ` `for` `(` `int` `i = ` `0` `; i < N; i++) ` ` ` `{` ` ` ` ` `// If the character is uppercase` ` ` `if` `(S.charAt(i) >= ` `'A'` `&& S.charAt(i) <= ` `'Z'` `)` ` ` `{` ` ` `currentSum++;` ` ` `}` ` ` `// Otherwise` ` ` `else` ` ` `currentSum--;` ` ` `// If currsum is 0` ` ` `if` `(currentSum == ` `0` `)` ` ` `res++;` ` ` `// If the current sum exists in` ` ` `// the HashMap prevSum` ` ` `if` `(prevSum.containsKey(currentSum))` ` ` `{` ` ` ` ` `// Increment the resultant` ` ` `// count by 1` ` ` `res += prevSum.get(currentSum);` ` ` `}` ` ` `// Update the frequency of the` ` ` `// current sum by 1` ` ` `prevSum.put(currentSum,` ` ` `prevSum.getOrDefault(currentSum, ` `0` `) + ` `1` `);` ` ` `}` ` ` `// Return the resultant count of` ` ` `// the subarrays` ` ` `return` `res;` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `String S = ` `"gEEk"` `;` ` ` `System.out.println(countSubstring(S, S.length()));` `}` `}` `// This code is contributed by abhinavjain194` |

## Python3

`# Python3 program for the above approach` `# Function to find the count of` `# substrings having an equal number` `# of uppercase and lowercase characters` `def` `countSubstring(S, N):` ` ` ` ` `# Stores the count of prefixes` ` ` `# having sum S considering uppercase` ` ` `# and lowercase characters as 1 and -1` ` ` `prevSum ` `=` `{}` ` ` `# Stores the count of substrings` ` ` `# having equal number of lowercase` ` ` `# and uppercase characters` ` ` `res ` `=` `0` ` ` `# Stores the sum obtained so far` ` ` `currentSum ` `=` `0` ` ` `for` `i ` `in` `range` `(N):` ` ` ` ` `# If the character is uppercase` ` ` `if` `(S[i] >` `=` `'A'` `and` `S[i] <` `=` `'Z'` `):` ` ` `currentSum ` `+` `=` `1` ` ` `# Otherwise` ` ` `else` `:` ` ` `currentSum ` `-` `=` `1` ` ` `# If currsum is o` ` ` `if` `(currentSum ` `=` `=` `0` `):` ` ` `res ` `+` `=` `1` ` ` `# If the current sum exists in` ` ` `# the HashMap prevSum` ` ` `if` `(currentSum ` `in` `prevSum):` ` ` ` ` `# Increment the resultant` ` ` `# count by 1` ` ` `res ` `+` `=` `(prevSum[currentSum])` ` ` `# Update the frequency of the` ` ` `# current sum by 1` ` ` `if` `currentSum ` `in` `prevSum:` ` ` `prevSum[currentSum] ` `+` `=` `1` ` ` `else` `:` ` ` `prevSum[currentSum] ` `=` `1` ` ` `# Return the resultant count of` ` ` `# the subarrays` ` ` `return` `res` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `S ` `=` `"gEEk"` ` ` `print` `(countSubstring(S, ` `len` `(S)))` `# This code is contributed by bgangwar59` |

## C#

`// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG{` `// Function to find the count of` `// substrings having an equal number` `// of uppercase and lowercase characters` `static` `int` `countSubstring(String S, ` `int` `N)` `{` ` ` ` ` `// Stores the count of prefixes` ` ` `// having sum S considering uppercase` ` ` `// and lowercase characters as 1 and -1` ` ` `Dictionary<` `int` `, ` ` ` `int` `> prevSum = ` `new` `Dictionary<` `int` `, ` ` ` `int` `>();` ` ` `// Stores the count of substrings` ` ` `// having equal number of lowercase` ` ` `// and uppercase characters` ` ` `int` `res = 0;` ` ` `// Stores the sum obtained so far` ` ` `int` `currentSum = 0;` ` ` `for` `(` `int` `i = 0; i < N; i++) ` ` ` `{` ` ` ` ` `// If the character is uppercase` ` ` `if` `(S[i] >= ` `'A'` `&& S[i] <= ` `'Z'` `)` ` ` `{` ` ` `currentSum++;` ` ` `}` ` ` `// Otherwise` ` ` `else` ` ` `currentSum--;` ` ` `// If currsum is 0` ` ` `if` `(currentSum == 0)` ` ` `res++;` ` ` `// If the current sum exists in` ` ` `// the Dictionary prevSum` ` ` `if` `(prevSum.ContainsKey(currentSum))` ` ` `{` ` ` ` ` `// Increment the resultant` ` ` `// count by 1` ` ` `res += prevSum[currentSum];` ` ` `prevSum[currentSum] = prevSum[currentSum] + 1;` ` ` `}` ` ` `else` ` ` `prevSum.Add(currentSum, 1);` ` ` `}` ` ` `// Return the resultant count of` ` ` `// the subarrays` ` ` `return` `res;` `}` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` `String S = ` `"gEEk"` `;` ` ` `Console.WriteLine(countSubstring(S, S.Length));` `}` `}` `// This code is contributed by Princi Singh` |

## Javascript

`<script>` `// Javascript program for the above approach` `// Function to find the count of` `// substrings having an equal number` `// of uppercase and lowercase characters` `function` `countSubstring(S, N)` `{` ` ` `// Stores the count of prefixes` ` ` `// having sum S considering uppercase` ` ` `// and lowercase characters as 1 and -1` ` ` `var` `prevSum = ` `new` `Map();` ` ` `// Stores the count of substrings` ` ` `// having equal number of lowercase` ` ` `// and uppercase characters` ` ` `var` `res = 0;` ` ` `// Stores the sum obtained so far` ` ` `var` `currentSum = 0;` ` ` `for` `(` `var` `i = 0; i < N; i++) {` ` ` `// If the character is uppercase` ` ` `if` `(S[i] >= ` `'A'` `&& S[i] <= ` `'Z'` `) {` ` ` `currentSum++;` ` ` `}` ` ` `// Otherwise` ` ` `else` ` ` `currentSum--;` ` ` `// If currsum is o` ` ` `if` `(currentSum == 0)` ` ` `res++;` ` ` `// If the current sum exists in` ` ` `// the HashMap prevSum` ` ` `if` `(prevSum.has(currentSum)) {` ` ` `// Increment the resultant` ` ` `// count by 1` ` ` `res += (prevSum.get(currentSum));` ` ` `}` ` ` `// Update the frequency of the` ` ` `// current sum by 1` ` ` `if` `(prevSum.has(currentSum))` ` ` `prevSum.set(currentSum, prevSum.get(currentSum)+1)` ` ` `else` ` ` `prevSum.set(currentSum, 1)` ` ` `}` ` ` `// Return the resultant count of` ` ` `// the subarrays` ` ` `return` `res;` `}` `// Driver Code` `var` `S = ` `"gEEk"` `;` `document.write( countSubstring(S, S.length));` `</script>` |

**Output:**

3

**Time Complexity:** O(N)**Auxiliary Space:** O(1)

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