Linear Search Algorithm

Author: OmkarPathak

CompetitiveProgramming/HackerEarth/Algorithms/Searching/P06_RankIt.py

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+# You have been given an array A consisting of N integers. All the elements in this array A are unique. You have to
+# answer some queries based on the elements of this array. Each query will consist of a single integer x. You need to
+# print the rank based position of this element in this array considering that the array is 1 indexed. The rank
+# based position of an element in an array is its position in the array when the array has been sorted in ascending order.
+#
+# Note: It is guaranteed that all the elements in this array are unique and for each x belonging to a query, value ′x′
+# shall exist in the array
+#
+# Input Format
+#
+# The first line consists of a single integer N denoting the size of array A. The next line contains N unique integers,
+# denoting the content of array A. The next line contains a single integer q denoting the number of queries. Each of
+# the next q lines contains a single integer x denoting the element whose rank based position needs to be printed.
+#
+# Output Format
+#
+# You need to print q integers denoting the answer to each query.
+#
+# Constraints
+#
+# 1≤N≤105
+# 1≤A[i]≤109
+# 1≤q≤105
+# 1≤x≤109
+#
+# SAMPLE INPUT
+# 5
+# 1 2 3 4 5
+# 5
+# 1
+# 2
+# 3
+# 4
+# 5
+#
+# SAMPLE OUTPUT
+# 1
+# 2
+# 3
+# 4
+# 5
+
+n = int(input())
+array = [int(i) for i in input().split()]
+array.insert(0, 0)
+array.sort()
+q = int(input())
+
+def binarySearch(low, high, element):
+  while(low <= high):
+    mid = (low + high) // 2
+    if array[mid] == element:
+      return mid
+    elif array[mid] < element:
+      low = mid + 1
+    else:
+      high = mid - 1
+
+for i in range(q):
+  number = int(input())
+  print(binarySearch(0,len(array), number))