Completed binary search1

.vscode/tasks.json

@@ -0,0 +1,29 @@
+{
+    "tasks": [
+        {
+            "type": "cppbuild",
+            "label": "C/C++: clang build active file",
+            "command": "/usr/bin/clang",
+            "args": [
+                "-fcolor-diagnostics",
+                "-fansi-escape-codes",
+                "-g",
+                "${file}",
+                "-o",
+                "${fileDirname}/${fileBasenameNoExtension}"
+            ],
+            "options": {
+                "cwd": "${fileDirname}"
+            },
+            "problemMatcher": [
+                "$gcc"
+            ],
+            "group": {
+                "kind": "build",
+                "isDefault": true
+            },
+            "detail": "Task generated by Debugger."
+        }
+    ],
+    "version": "2.0.0"
+}

Problem1.cpp

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+// Time Complexity : O(log(m * n)
+// Space Complexity : O(1)
+// 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 in three sentences only
+// Problem1 Search a 2D Matrix(https://leetcode.com/problems/search-a-2d-matrix/)
+
+#include <iostream>
+#include <vector>
+using namespace std;
+
+class Solution {
+public:
+    bool searchMatrix(vector<vector<int>>& matrix, int target) {
+        int m = matrix.size();
+        if (m == 0) return false;
+        int n = matrix[0].size();
+
+        //get start and end of the matrix
+        int start = 0;
+        int end = m * n - 1;
+
+        //for binary search get mid value
+        while (start <= end) {
+            int mid = start + (end - start) / 2;
+            int midVal = matrix[mid / n][mid % n];
+
+            //check if midval is equal to target or else, compare it to decide which side of matrix to move
+            if (midVal == target) {
+                return true;
+            } else if (target < midVal) {
+                end = mid - 1; 
+            } else {
+                start = mid + 1; 
+            }
+        }
+// if target isnt found return false
+        return false;
+    }
+};
+

problem2.cpp

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+// Time Complexity : O(log n)
+// Space Complexity : O(1)
+// 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 in three sentences only
+
+#include <iostream>
+#include <vector>
+
+using namespace std;
+
+class Solution {
+public:
+    int search(vector<int>& nums, int target) {
+        int start = 0;
+        int end = nums.size() - 1;
+
+        //to perform binary search on one part of the array. in a rotated sorted array one half is sorted
+        while (start <= end) {
+            int mid = start + (end - start) / 2;
+
+            if (nums[mid] == target)
+                return mid;
+
+            // Check if left half is sorted
+            if (nums[start] <= nums[mid]) {
+                if (nums[start] <= target && target < nums[mid])
+                    end = mid - 1;
+                else
+                    start = mid + 1;
+            }
+            // if Right half is sorted
+            else {
+                if (nums[mid] < target && target <= nums[end])
+                    start = mid + 1;
+                else
+                    end = mid - 1;
+            }
+        }
+// if target isnt found
+        return -1;
+    }
+};

problem3.cpp

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+// Time Complexity : O(log k)
+// Space Complexity : O(1)
+// 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 in three sentences only
+
+//Search in Infinite sorted array: 
+
+//https://leetcode.com/problems/search-in-a-sorted-array-of-unknown-size/
+
+//Given a sorted array of unknown length and a number to search for, return the index of the number in the array. Accessing an element out of bounds throws exception. If the number occurs multiple times, return the index of any occurrence. If it isn’t present, return -1.
+
+#include <climits>
+
+class ArrayReader {
+    public:
+    int get(int index);
+};
+
+class Solution {
+    public:
+    int search(const ArrayReader& reader, int target) {
+        int start = 0, end = 1;
+
+        while(reader.get(end) < target) {
+            start = end;
+            end *=2;
+        }
+
+        while(reader.get(end) < target) {
+            start = end;
+            end *= 2;
+        }
+        while (start<=end) {
+            int mid = start + (end - start)/2;
+            int midVal = reader.get(mid);
+
+            if (midVal == target) return mid;
+            if (midVal > target || midVal == INT_MAX)
+            end = mid -1;
+            else
+            start = mid +1;
+        }
+        return -1;
+    }
+};