Binary TreesBinary Trees36

  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Postorder Traversal of a Binary Tree Using Recursion
  4. 4Postorder Traversal of a Binary Tree using Iteration
  5. 5Level Order Traversal of a Binary Tree using Recursion
  6. 6Level Order Traversal of a Binary Tree using Iteration
  7. 7Reverse Level Order Traversal of a Binary Tree using Iteration
  8. 8Reverse Level Order Traversal of a Binary Tree using Recursion
  9. 9Find Height of a Binary Tree
  10. 10Find Diameter of a Binary Tree
  11. 11Find Mirror of a Binary Tree - Todo
  12. 12Inorder Traversal of a Binary Tree using Recursion
  13. 13Inorder Traversal of a Binary Tree using Iteration
  14. 14Left View of a Binary Tree
  15. 15Right View of a Binary Tree
  16. 16Top View of a Binary Tree
  17. 17Bottom View of a Binary Tree
  18. 18Zigzag Traversal of a Binary Tree
  19. 19Check if a Binary Tree is Balanced
  20. 20Diagonal Traversal of a Binary Tree
  21. 21Boundary Traversal of a Binary Tree
  22. 22Construct a Binary Tree from a String with Bracket Representation
  23. 23Convert a Binary Tree into a Doubly Linked List
  24. 24Convert a Binary Tree into a Sum Tree
  25. 25Find Minimum Swaps Required to Convert a Binary Tree into a BST
  26. 26Check if a Binary Tree is a Sum Tree
  27. 27Check if All Leaf Nodes are at the Same Level in a Binary Tree
  28. 28Lowest Common Ancestor (LCA) in a Binary Tree
  29. 29Solve the Tree Isomorphism Problem
  30. 30Check if a Binary Tree Contains Duplicate Subtrees of Size 2 or More
  31. 31Check if Two Binary Trees are Mirror Images
  32. 32Calculate the Sum of Nodes on the Longest Path from Root to Leaf in a Binary Tree
  33. 33Print All Paths in a Binary Tree with a Given Sum
  34. 34Find the Distance Between Two Nodes in a Binary Tree
  35. 35Find the kth Ancestor of a Node in a Binary Tree
  36. 36Find All Duplicate Subtrees in a Binary Tree

Sum of Beauty of All Substrings



Problem Statement

Given a string s, your task is to find the sum of beauty for all its substrings.

The beauty of a substring is defined as the difference between the highest frequency character and the lowest frequency character (ignoring characters that do not appear).

For example, in substring "aabcb", the frequencies are: a → 2, b → 2, c → 1. So the beauty is 2 - 1 = 1.

You need to compute this value for every possible substring of s and return the total sum of all these beauties.

Examples

Input StringOutputDescription
"aabcb"5Each substring is checked for character frequency difference. Total beauty = 5
"abcab"5Beauty is computed for substrings like "abc", "bca", etc.
"aaaa"0All characters are the same in every substring, so beauty = 0
"ab"0Each substring has unique characters or equal frequencies
"a"0Only one character. No frequency difference
""0Empty string has no substrings. Answer is 0

Solution

To solve this problem, we need to understand what beauty means for a substring. The beauty is the difference between how often the most frequent character appears and how often the least frequent (but present) character appears in that substring.

We consider every possible substring of the input string. For each substring, we count how many times each character appears. We ignore characters that have a count of 0, and then find the difference between the highest and lowest frequency among the characters that do appear.

Let’s understand this through different types of input:

  • Case 1 – Normal string (e.g., "aabcb"):
    There are substrings where characters appear with different frequencies. For example, in "aab", 'a' appears twice, 'b' once → beauty = 1. We do this for every substring and keep adding up their beauty values.
  • Case 2 – All same characters (e.g., "aaaa"):
    Every substring has all characters the same, so the highest and lowest frequencies are equal. Beauty = 0 for all of them.
  • Case 3 – Short strings like "ab" or "a":
    In "ab", both characters appear once, so no frequency difference → beauty = 0. In "a", there's only one character → beauty = 0.
  • Case 4 – Empty string ""
    There are no substrings to evaluate, so the total beauty is 0.

This problem involves many substrings, but we can optimize the solution by using a frequency array (of size 26 for each character) and reusing it while expanding the substring window. In each iteration, we quickly find the max and min (non-zero) frequencies and calculate the beauty.

By combining smart counting with substring expansion, we can solve this efficiently even for longer strings. The time complexity is roughly O(n²) but works well with optimization.

Algorithm Steps

  1. Initialize a variable totalBeauty = 0.
  2. Loop over each start index i of the string.
  3. Initialize a frequency array count[26] for characters 'a' to 'z'.
  4. For each end index ji, do:
  5. → Increment the count of character s[j].
  6. → Find maxFreq and minFreq from the count array (ignoring 0s).
  7. → Add (maxFreq - minFreq) to totalBeauty.
  8. Return totalBeauty.

Code

Python
Java
JavaScript
C
C++
C#
Kotlin
Swift
Go
Php
def beautySum(s):
    total_beauty = 0
    n = len(s)

    for i in range(n):
        count = [0] * 26
        for j in range(i, n):
            idx = ord(s[j]) - ord('a')
            count[idx] += 1

            freq_list = [c for c in count if c > 0]
            max_freq = max(freq_list)
            min_freq = min(freq_list)

            total_beauty += (max_freq - min_freq)

    return total_beauty

# Example
s = "aabcb"
print("Sum of Beauty:", beautySum(s))

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(n^2)We iterate through all possible substrings (n^2 combinations), and each frequency check takes constant time.
Average CaseO(n^2)The main loop processes all substrings, and frequency difference is computed in O(26) = O(1).
Average CaseO(n^2)Even if all characters are distinct, the double loop covers all substrings of the input string.

Space Complexity

O(1)

Explanation: Only a fixed-size array of 26 elements is used for frequency counting.



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