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DSA Visualizations /

Overview of TechniquesOverview of Techniques14
❯

  1. 1Overview of DSA Problem Solving Techniques
  2. 2Brute Force Technique in DSA
  3. 3Greedy Algorithm Technique in DSA
  4. 4Divide and Conquer Technique in DSA
  5. 5Dynamic Programming Technique in DSA
  6. 6Backtracking Technique in DSA
  7. 7Recursion Technique in DSA
  8. 8Sliding Window Technique in DSA
  9. 9Two Pointers Technique
  10. 10Binary Search Technique
  11. 11Tree / Graph Traversal Technique in DSA
  12. 12Bit Manipulation Technique in DSA
  13. 13Hashing Technique
  14. 14Heaps Technique in DSA

SortingSorting10
❯

  1. 1Bubble Sort
  2. 2Selection Sort
  3. 3Insertion Sort
  4. 4Merge Sort
  5. 5Quick Sort
  6. 6Heap Sort
  7. 7Radix Sort
  8. 8Counting Sort
  9. 9Bucket Sort
  10. 10Shell Sort

ArraysArrays32
❯

  1. 1Find Maximum and Minimum in Array using Loop
  2. 2Find Second Largest in Array
  3. 3Find Second Smallest in Array
  4. 4Reverse Array using Two Pointers
  5. 5Check if Array is Sorted
  6. 6Remove Duplicates from Sorted Array
  7. 7Left Rotate an Array by One Place
  8. 8Left Rotate an Array by K Places
  9. 9Move Zeroes in Array to End
  10. 10Linear Search in Array
  11. 11Union of Two Arrays
  12. 12Find Missing Number in Array
  13. 13Max Consecutive Ones in Array
  14. 14Find Kth Smallest Element
  15. 15Longest Subarray with Given Sum (Positives)
  16. 16Longest Subarray with Given Sum (Positives and Negatives)
  17. 17Find Majority Element in Array (more than n/2 times)
  18. 18Find Majority Element in Array (more than n/3 times)
  19. 19Maximum Subarray Sum using Kadane's Algorithm
  20. 20Print Subarray with Maximum Sum
  21. 21Stock Buy and Sell
  22. 22Rearrange Array Alternating Positive and Negative Elements
  23. 23Next Permutation of Array
  24. 24Leaders in an Array
  25. 25Longest Consecutive Sequence in Array
  26. 26Count Subarrays with Given Sum
  27. 27Sort an Array of 0s, 1s, and 2s
  28. 28Two Sum Problem
  29. 29Three Sum Problem
  30. 304 Sum Problem
  31. 31Find Length of Largest Subarray with 0 Sum
  32. 32Find Maximum Product Subarray

Arrays - Binary SearchArrays - Binary Search28
❯

  1. 1Binary Search in Array using Iteration
  2. 2Find Lower Bound in Sorted Array
  3. 3Find Upper Bound in Sorted Array
  4. 4Search Insert Position in Sorted Array (Lower Bound Approach)
  5. 5Floor and Ceil in Sorted Array
  6. 6First Occurrence in a Sorted Array
  7. 7Last Occurrence in a Sorted Array
  8. 8Count Occurrences in Sorted Array
  9. 9Search Element in a Rotated Sorted Array
  10. 10Search in Rotated Sorted Array with Duplicates
  11. 11Minimum in Rotated Sorted Array
  12. 12Find Rotation Count in Sorted Array
  13. 13Search Single Element in Sorted Array
  14. 14Find Peak Element in Array
  15. 15Square Root using Binary Search
  16. 16Nth Root of a Number using Binary Search
  17. 17Koko Eating Bananas
  18. 18Minimum Days to Make M Bouquets
  19. 19Find the Smallest Divisor Given a Threshold
  20. 20Capacity to Ship Packages within D Days
  21. 21Kth Missing Positive Number
  22. 22Aggressive Cows Problem
  23. 23Allocate Minimum Number of Pages
  24. 24Split Array - Minimize Largest Sum
  25. 25Painter's Partition Problem
  26. 26Minimize Maximum Distance Between Gas Stations
  27. 27Median of Two Sorted Arrays of Different Sizes
  28. 28K-th Element of Two Sorted Arrays

StringsStrings18
❯

  1. 1Reverse Words in a String
  2. 2Find the Largest Odd Number in a Numeric String
  3. 3Find Longest Common Prefix in Array of Strings
  4. 4Check If Two Strings Are Isomorphic - Optimal HashMap Solution
  5. 5Check String Rotation using Concatenation - Optimal Approach
  6. 6Check if Two Strings Are Anagrams - Optimal Approach
  7. 7Sort Characters by Frequency - Optimal HashMap and Heap Approach
  8. 8Find Longest Palindromic Substring - Dynamic Programming Approach
  9. 9Find Longest Palindromic Substring Without Dynamic Programming
  10. 10Remove Outermost Parentheses in String
  11. 11Find Maximum Nesting Depth of Parentheses - Optimal Stack-Free Solution
  12. 12Convert Roman Numerals to Integer - Efficient Approach
  13. 13Convert Integer to Roman Numeral - Step-by-Step for Beginners
  14. 14Implement Atoi - Convert String to Integer in Java
  15. 15Count Number of Substrings in a String - Explanation with Formula
  16. 16Edit Distance Problem
  17. 17Calculate Sum of Beauty of All Substrings - Optimal Approach
  18. 18Reverse Each Word in a String - Optimal Approach

MatrixMatrix3
❯

  1. 1Set Matrix Zeroes
  2. 2Rotate Matrix by 90 Degrees Clockwise
  3. 3Print Matrix in Spiral Manner

Matrix - Binary SearchMatrix - Binary Search5
❯

  1. 1Row with Maximum Number of 1's
  2. 2Search in a Sorted 2D Matrix
  3. 3Search in Row and Column-wise Sorted Matrix
  4. 4Find Target in 2D Matrix
  5. 5Matrix Median

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

Binary Search TreesBinary Search Trees14
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  1. 1Find a Value in a Binary Search Tree
  2. 2Delete a Node in a Binary Search Tree
  3. 3Find the Minimum Value in a Binary Search Tree
  4. 4Find the Maximum Value in a Binary Search Tree
  5. 5Find the Inorder Successor in a Binary Search Tree
  6. 6Find the Inorder Predecessor in a Binary Search Tree
  7. 7Check if a Binary Tree is a Binary Search Tree
  8. 8Find the Lowest Common Ancestor of Two Nodes in a Binary Search Tree
  9. 9Convert a Binary Tree into a Binary Search Tree
  10. 10Balance a Binary Search Tree
  11. 11Merge Two Binary Search Trees
  12. 12Find the kth Largest Element in a Binary Search Tree
  13. 13Find the kth Smallest Element in a Binary Search Tree
  14. 14Flatten a Binary Search Tree into a Sorted List

Aggressive Cows Problem
Optimal Binary Search Solution

Topic Contents

ProblemExamplesSolutionVisualizationAlgorithmCodeTimeSpace

Problem Statement

Given N stalls represented by their positions on a number line (sorted or unsorted), and K cows to place in these stalls, the goal is to place the cows such that the minimum distance between any two cows is as large as possible.

If the number of cows exceeds the number of stalls, it’s not possible to place all cows.

Examples❯

Stall PositionsNo. of Cows (K)OutputDescription
[1, 2, 4, 8, 9]33We can place cows at positions 1, 4, and 8 — min distance = 3
[1, 2, 4, 8, 9]28Place cows at positions 1 and 9 — max possible distance
[5, 17, 100, 111]2106Place cows at 5 and 111 — max min distance is 106
[5, 17, 100, 111]46All cows must be placed, closest pair ends up at 6 units apart
[10]10Only one cow, no distance to maximize
[10]2-1Not enough stalls to place 2 cows
[]2-1Empty array, no stalls available
[2, 4, 6]00Zero cows, so minimum distance is trivially zero

Solution❯

The Aggressive Cows problem is a classic example where we want to use binary search on the answer. Instead of searching through positions, we ask: what is the largest possible minimum distance we can achieve between cows placed in stalls?

Understanding the Problem

We are given stall positions — say [1, 2, 4, 8, 9] — and asked to place K cows in them. The twist is: we want to maximize the minimum distance between any two cows. That means, the cows should be as far apart as possible, but we still need to fit all of them into the stalls.

Different Scenarios to Consider

  • Exact fit: If the number of cows equals the number of stalls, the only option is one cow per stall, and the smallest distance between any two is what we return.
  • Single cow: If there’s only one cow to place, we can put it anywhere. There’s no pair to measure distance, so the answer is 0.
  • Too many cows: If K > N, we cannot place all cows — so we return -1 or some invalid flag.
  • Empty stalls: If the stall list is empty, there’s no way to place any cows — again, return -1.

How the Strategy Works

First, we sort the stall positions to ensure they're in order. Then, we try to guess the minimum distance between cows (starting from 1 up to the max difference between stalls) using binary search.

At each step, we check: "Can we place all K cows such that the gap between any two is at least mid?"

  • If yes → maybe we can do even better, so try a bigger distance.
  • If no → that distance is too much, reduce the range.

We keep narrowing the range until we find the largest distance that works — that’s our answer.

Efficiency

This method is very efficient: we’re doing a binary search on distance (which ranges from 1 to max stall gap), and for each guess, we do a linear pass to see if placement is possible. This gives us an overall time complexity of O(N log(MaxDistance)).

It’s an excellent real-world inspired problem that blends greedy thinking with binary search — ideal for learning how to "search on the answer".

Visualization

Algorithm Steps

  1. Sort the arr in increasing order.
  2. Set low = 1 (smallest possible min distance) and high = arr[n - 1] - arr[0] (largest possible).
  3. Use binary search: while low ≤ high:
  4. → Compute mid = (low + high) / 2.
  5. → Check if it's possible to place k cows with at least mid distance between them.
  6. → If yes, store mid as a candidate and try for a larger value (low = mid + 1).
  7. → If not, reduce the distance (high = mid - 1).
  8. Return the largest valid mid found.

Code

Java
Python
JavaScript
C
C++
C#
Kotlin
Swift
Go
Php
import java.util.Arrays;

public class AggressiveCows {
    public static boolean canPlace(int[] stalls, int cows, int minDist) {
        int count = 1; // Place first cow at the first stall
        int lastPlaced = stalls[0];
        for (int i = 1; i < stalls.length; i++) {
            if (stalls[i] - lastPlaced >= minDist) {
                count++;
                lastPlaced = stalls[i];
                if (count == cows) return true;
            }
        }
        return false;
    }

    public static int solve(int n, int k, int[] arr) {
        Arrays.sort(arr);
        int low = 1, high = arr[n - 1] - arr[0];
        int result = 0;

        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (canPlace(arr, k, mid)) {
                result = mid; // Try for a bigger answer
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        return result;
    }

    public static void main(String[] args) {
        int[] stalls = {1, 2, 4, 8, 9};
        int k = 3;
        System.out.println("Max Min Distance: " + solve(stalls.length, k, stalls));
    }
}

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(n log(maxDist))Binary search on the distance (log(max position)) and each check takes O(n).
Average CaseO(n log(maxDist))Each iteration of binary search checks placement for given mid using O(n).
Average CaseO(n log(maxDist))In the worst case, binary search runs log(max distance) steps and each step scans the stalls.

Space Complexity

O(1)

Explanation: No extra space apart from a few variables; we sort in-place and use constant memory.

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