Competative Coding 3

PascalTraingle.java

@@ -0,0 +1,40 @@
+// Time Complexity :O(numRows^2)
+// Space Complexity :O(1) if we don't consider the output list, otherwise O(numRows^2) for the output list
+// Did this code successfully run on Leetcode :Yes
+// Any problem you faced while coding this :No , a bit of confusion in the inner loop but I got it after dry run
+
+// Your code here along with comments explaining your approach
+//The first and last elements of every row are always 1
+//Every middle element is the sum of the two elements directly above it from the previous row
+//So, if we already know the previous row, we can build the current row easily by applying this rule.
+//If it's the first or last position, add 1
+//Otherwise, compute the value as:
+//res[i-1][j-1] + res[i-1][j] (from the previous row)
+import java.util.*;
+
+class Solution {
+    public List<List<Integer>> generate(int numRows) {
+
+        List<List<Integer>> res = new ArrayList<>();
+
+        res.add(new ArrayList<>(Arrays.asList(1)));
+
+        for (int i = 1; i < numRows; i++) {
+
+            List<Integer> list = new ArrayList<>();
+
+            for (int j = 0; j <= i; j++) {
+
+                if (j == 0 || j == i) {
+                    list.add(1);
+                } else {
+                    list.add(res.get(i - 1).get(j - 1) + res.get(i - 1).get(j));
+                }
+            }
+
+            res.add(list);
+        }
+
+        return res;
+    }
+}

SumofK.java

@@ -0,0 +1,43 @@
+// Time Complexity :    o(n) where n is the number of elements in the input array
+// Space Complexity : O(n) where n is the number of unique elements in the input array
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No
+
+
+// Your code here along with comments explaining your approach
+// We can use a HashMap to store the frequency of each number in the input array as we need uniqy pairs and
+//  we can easily check if the pair exists in the map
+// We can iterate through the map and check for each key if the pair (key + k) exists in the map
+// If k is 0, we need to check if the frequency of the key is greater than 1 to count it as a valid pair,
+//  otherwise we can simply check if the pair exists
+// We can keep a count of valid pairs and return it at the end
+import java.util.*;
+
+class Solution {
+    public int findPairs(int[] nums, int k) {
+
+        Map<Integer, Integer> map = new HashMap<>();
+        int count = 0;
+
+        // build frequency map
+        for (int num : nums) {
+            map.put(num, map.getOrDefault(num, 0) + 1);
+        }
+
+        // check pairs
+        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
+
+            int key = entry.getKey();
+            int value = entry.getValue();
+
+            if (k == 0 && value > 1) {
+                count++;
+            }
+            else if (k != 0 && map.containsKey(key + k)) {
+                count++;
+            }
+        }
+
+        return count;
+    }
+}