add first and fast position of element in sorted Array

Author: akgmage

Binary_Search/first_and_last_pos_of_element.cpp

@@ -0,0 +1,70 @@
+/*
+Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value.
+
+If target is not found in the array, return [-1, -1].
+
+You must write an algorithm with O(log n) runtime complexity.
+
+ 
+
+Example 1:
+
+Input: nums = [5,7,7,8,8,10], target = 8
+Output: [3,4]
+Example 2:
+
+Input: nums = [5,7,7,8,8,10], target = 6
+Output: [-1,-1]
+Example 3:
+
+Input: nums = [], target = 0
+Output: [-1,-1]
+ 
+
+Constraints:
+
+0 <= nums.length <= 105
+-109 <= nums[i] <= 109
+nums is a non-decreasing array.
+-109 <= target <= 109
+
+*/
+#include<bits/stdc++.h>
+class Solution {
+public:
+    int get_index(vector<int>& nums, int target, bool found){
+        int ans = -1;
+        int start = 0, end = nums.size() - 1;
+        while(start <= end){
+            int mid = start + (end - start) / 2;
+            if(nums[mid] == target){
+                ans = mid;
+                if(found){
+                    end = mid - 1; // search in left part
+                }
+                else{
+                    start = mid + 1; // search in right part
+                }
+            }
+            else if(nums[mid] > target){
+                end = mid - 1;
+            }
+            else{
+                start = mid + 1;
+            }
+        }
+        return ans;
+    }
+    vector<int> searchRange(vector<int>& nums, int target) {
+        vector<int> ans(2, -1);
+        int first = get_index(nums, target, true);
+        if(first == -1)
+            return ans;
+        int last = get_index(nums, target, false);
+        //ans.push_back(first);
+        //ans.push_back(last);
+        ans[0] = first;
+        ans[1] = last;
+        return ans;
+    }
+};