Binary TreesBinary Trees36
  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Inorder Traversal of a Binary Tree using Recursion
  4. 4Inorder Traversal of a Binary Tree using Iteration
  5. 5Postorder Traversal of a Binary Tree Using Recursion
  6. 6Postorder Traversal of a Binary Tree using Iteration
  7. 7Level Order Traversal of a Binary Tree using Recursion
  8. 8Level Order Traversal of a Binary Tree using Iteration
  9. 9Reverse Level Order Traversal of a Binary Tree using Iteration
  10. 10Reverse Level Order Traversal of a Binary Tree using Recursion
  11. 11Find Height of a Binary Tree
  12. 12Find Diameter of a Binary Tree
  13. 13Find Mirror of a Binary Tree
  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
GraphsGraphs46
  1. 1Breadth-First Search in Graphs
  2. 2Depth-First Search in Graphs
  3. 3Number of Provinces in an Undirected Graph
  4. 4Connected Components in a Matrix
  5. 5Rotten Oranges Problem - BFS in Matrix
  6. 6Flood Fill Algorithm - Graph Based
  7. 7Detect Cycle in an Undirected Graph using DFS
  8. 8Detect Cycle in an Undirected Graph using BFS
  9. 9Distance of Nearest Cell Having 1 - Grid BFS
  10. 10Surrounded Regions in Matrix using Graph Traversal
  11. 11Number of Enclaves in Grid
  12. 12Word Ladder - Shortest Transformation using Graph
  13. 13Word Ladder II - All Shortest Transformation Sequences
  14. 14Number of Distinct Islands using DFS
  15. 15Check if a Graph is Bipartite using DFS
  16. 16Topological Sort Using DFS
  17. 17Topological Sort using Kahn's Algorithm
  18. 18Cycle Detection in Directed Graph using BFS
  19. 19Course Schedule - Task Ordering with Prerequisites
  20. 20Course Schedule 2 - Task Ordering Using Topological Sort
  21. 21Find Eventual Safe States in a Directed Graph
  22. 22Alien Dictionary Character Order
  23. 23Shortest Path in Undirected Graph with Unit Distance
  24. 24Shortest Path in DAG using Topological Sort
  25. 25Dijkstra's Algorithm Using Set - Shortest Path in Graph
  26. 26Dijkstra’s Algorithm Using Priority Queue
  27. 27Shortest Distance in a Binary Maze using BFS
  28. 28Path With Minimum Effort in Grid using Graphs
  29. 29Cheapest Flights Within K Stops - Graph Problem
  30. 30Number of Ways to Reach Destination in Shortest Time - Graph Problem
  31. 31Minimum Multiplications to Reach End - Graph BFS
  32. 32Bellman-Ford Algorithm for Shortest Paths
  33. 33Floyd Warshall Algorithm for All-Pairs Shortest Path
  34. 34Find the City With the Fewest Reachable Neighbours
  35. 35Minimum Spanning Tree in Graphs
  36. 36Prim's Algorithm for Minimum Spanning Tree
  37. 37Disjoint Set (Union-Find) with Union by Rank and Path Compression
  38. 38Kruskal's Algorithm - Minimum Spanning Tree
  39. 39Minimum Operations to Make Network Connected
  40. 40Most Stones Removed with Same Row or Column
  41. 41Accounts Merge Problem using Disjoint Set Union
  42. 42Number of Islands II - Online Queries using DSU
  43. 43Making a Large Island Using DSU
  44. 44Bridges in Graph using Tarjan's Algorithm
  45. 45Articulation Points in Graphs
  46. 46Strongly Connected Components using Kosaraju's Algorithm

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