From 3ce37f1412bb3e476ff4ff0ba8dd31f84935cfb8 Mon Sep 17 00:00:00 2001
From: Harshitha Thondalapally <harshi@Harshithas-Laptop.local>
Date: Mon, 12 Jan 2026 12:00:01 -0500
Subject: [PATCH] Completed competitive coding 3

---
 src/KDiffPairs.java      | 85 ++++++++++++++++++++++++++++++++++++++++
 src/PascalsTriangle.java | 54 +++++++++++++++++++++++++
 2 files changed, 139 insertions(+)
 create mode 100644 src/KDiffPairs.java
 create mode 100644 src/PascalsTriangle.java

diff --git a/src/KDiffPairs.java b/src/KDiffPairs.java
new file mode 100644
index 00000000..b058c560
--- /dev/null
+++ b/src/KDiffPairs.java
@@ -0,0 +1,85 @@
+/*
+Problem:
+Given an array of integers nums and an integer k, return the number of UNIQUE k-diff pairs in the array.
+A k-diff pair is (a, b) such that |a - b| = k.
+
+Key points:
+- Pairs are unique by VALUE, not by index.
+- If k < 0, answer is 0 (because absolute difference can't be negative).
+
+Approach (HashMap frequency):
+1) Build a frequency map: freq[x] = how many times x appears.
+2) Iterate over each unique value "key" in the map:
+   - If k == 0:
+       We need pairs like (key, key). This is only valid if freq[key] >= 2.
+   - If k > 0:
+       We only count (key, key + k) to avoid double counting.
+       If (key + k) exists in the map, we count 1 unique pair.
+3) Return count.
+Why no double counting for k > 0?
+- We only look in one direction: key -> key + k.
+  So (a, b) is counted once, not again as (b, a).
+Time Complexity:
+ O(n) average time
+Space Complexity:
+- worst case O(n)
+*/
+
+import java.util.HashMap;
+import java.util.Arrays;
+
+public class KDiffPairs {
+
+    public int findPairs(int[] nums, int k) {
+        // Absolute difference cannot be negative
+        if (k < 0) return 0;
+
+        HashMap<Integer, Integer> freq = new HashMap<>();
+        for (int x : nums) {
+            freq.put(x, freq.getOrDefault(x, 0) + 1);
+        }
+
+        int count = 0;
+
+        for (int key : freq.keySet()) {
+            if (k == 0) {
+                // Need at least two occurrences to form (key, key)
+                if (freq.get(key) > 1) count++;
+            } else {
+                // Count pair (key, key + k) only once
+                int complement = key + k;
+                if (freq.containsKey(complement)) count++;
+            }
+        }
+
+        return count;
+    }
+
+    public static void main(String[] args) {
+        KDiffPairs solver = new KDiffPairs();
+
+        int[][] testArrays = {
+                {3, 1, 4, 1, 5},   // k=2 => 2
+                {1, 2, 3, 4, 5},   // k=1 => 4
+                {1, 3, 1, 5, 4},   // k=0 => 1
+                {1, 1, 1, 1},      // k=0 => 1
+                {1, 1, 2, 2},      // k=1 => 1
+                {-1, -2, -3},      // k=1 => 2
+                {1, 2, 3}          // k=-1 => 0
+        };
+
+        int[] ks = {2, 1, 0, 0, 1, 1, -1};
+        int[] expected = {2, 4, 1, 1, 1, 2, 0};
+
+        for (int i = 0; i < testArrays.length; i++) {
+            int ans = solver.findPairs(testArrays[i], ks[i]);
+
+            System.out.println("Test " + (i + 1));
+            System.out.println("nums: " + Arrays.toString(testArrays[i]));
+            System.out.println("k: " + ks[i]);
+            System.out.println("output: " + ans);
+            System.out.println("expected: " + expected[i]);
+            System.out.println();
+        }
+    }
+}
diff --git a/src/PascalsTriangle.java b/src/PascalsTriangle.java
new file mode 100644
index 00000000..590e7cb4
--- /dev/null
+++ b/src/PascalsTriangle.java
@@ -0,0 +1,54 @@
+/*
+Problem: Pascal's Triangle
+Given an integer numRows, return the first numRows of Pascal’s triangle.
+
+Rules:
+- The first and last element of every row is 1.
+- Any inner element at (row i, col j) is the sum of the two elements above it:
+  triangle[i][j] = triangle[i-1][j-1] + triangle[i-1][j]
+
+Approach:
+1) Create an outer list to store all rows.
+2) For each row i from 0 to numRows-1:
+   - Create a list of size (i + 1)
+   - For each column j from 0 to i:
+       - If j == 0 or j == i, add 1 (edges)
+       - Else add result[i-1][j-1] + result[i-1][j] (from previous row)
+3) Return the triangle.
+
+Time Complexity: O(numRows^2)  (total elements in triangle = 1 + 2 + ... + numRows)
+Space Complexity: O(numRows^2) for storing the result (excluding output is not possible here since we must return it)
+*/
+
+import java.util.ArrayList;
+import java.util.List;
+
+public class PascalsTriangle {
+    public List<List<Integer>> generate(int numRows) {
+        List<List<Integer>> result = new ArrayList<List<Integer>>();
+        if(numRows == 0){
+            return result;
+        }
+        for(int i = 0; i < numRows; i++){
+            List<Integer> row = new ArrayList<>(i+1);
+            for(int j = 0; j <=i ; j++){
+                if(j == 0 || j == i){
+                    row.add(1);
+                }
+                else{
+                    List<Integer> temp = result.get(i-1);
+                    row.add(temp.get(j-1) + temp.get(j));
+                }
+            }
+            result.add(row);
+        }
+        return  result;
+    }
+    public static void main(String[] args) {
+        PascalsTriangle t = new PascalsTriangle();
+        List<List<Integer>> result = t.generate(5);
+        for(List<Integer> row : result){
+            System.out.println(row);
+        }
+    }
+}