Coding Hashing PYTHON Time Complexity

Python Programming – check for pair in A[] with sum as x

Time Complexity: Depends on what sorting algorithm we use. If we use Merge Sort or Heap Sort then (-)(nlogn) in worst case.

METHOD 1 (Use Sorting)


hasArrayTwoCandidates (A[], ar_size, sum)
1) Sort the array in non-decreasing order.
2) Initialize two index variables to find the candidate 
   elements in the sorted array.
       (a) Initialize first to the leftmost index: l = 0
       (b) Initialize second  the rightmost index:  r = ar_size-1
3) Loop while l < r.
       (a) If (A[l] + A[r] == sum)  then return 1
       (b) Else if( A[l] + A[r] <  sum )  then l++
       (c) Else r--    
4) No candidates in whole array - return 0

Time Complexity: Depends on what sorting algorithm we use. If we use Merge Sort or Heap Sort then (-)(nlogn) in worst case. If we use Quick Sort then O(n^2) in worst case.
Auxiliary Space : Again, depends on sorting algorithm. For example auxiliary space is O(n) for merge sort and O(1) for Heap Sort.

Let Array be {1, 4, 45, 6, 10, -8} and sum to find be 16

Sort the array
A = {-8, 1, 4, 6, 10, 45}

Initialize l = 0, r = 5
A[l] + A[r] ( -8 + 45) > 16 => decrement r. Now r = 10
A[l] + A[r] ( -8 + 10) < 2 => increment l. Now l = 1
A[l] + A[r] ( 1 + 10) < 16 => increment l. Now l = 2
A[l] + A[r] ( 4 + 10) < 14 => increment l. Now l = 3
A[l] + A[r] ( 6 + 10) == 16 => Found candidates (return 1)

Note: If there are more than one pair having the given sum then this algorithm reports only one. Can be easily extended for this though.


Python Program
# Python program to check for the sum condition to be satisified
def hasArrayTwoCandidates(A,arr_size,sum):
    # sort the array
    l = 0
    r = arr_size-1
    # traverse the array for the two elements
    while l<r:
        if (A[l] + A[r] == sum):
            return 1
        elif (A[l] + A[r] < sum):
            l += 1
            r -= 1
    return 0
# Implementation of Quick Sort
# A[] --> Array to be sorted
# si  --> Starting index
# ei  --> Ending index
def quickSort(A, si, ei):
    if si < ei:
# Utility function for partitioning the array(used in quick sort)
def partition(A, si, ei):
    x = A[ei]
    i = (si-1)
    for j in range(si,ei):
        if A[j] <= x:
            i += 1
            # This operation is used to swap two variables is python
            A[i], A[j] = A[j], A[i]
        A[i+1], A[ei] = A[ei], A[i+1]
    return i+1
# Driver program to test the functions
A = [1,4,45,6,10,-8]
n = 16
if (hasArrayTwoCandidates(A, len(A), n)):
    print("Array has two elements with the given sum")
    print("Array doesn't have two elements with the given sum")


Array has two elements with the given sum

METHOD 2 (Use Hash Map)
Thanks to Bindu for suggesting this method and thanks to Shekhu for providing code.
This method works in O(n) time if range of numbers is known.
Let sum be the given sum and A[] be the array in which we need to find pair.

1) Initialize Binary Hash Map M[] = {0, 0, ...}
2) Do following for each element A[i] in A[]
   (a)	If M[x - A[i]] is set then print the pair (A[i], x - A[i])
   (b)	Set M[A[i]]


Python Program
# Python program to find if there are two elements wtih given sum
CONST_MAX = 100000
# function to check for the given sum in the array
def printPairs(arr, arr_size, sum):
    # initialize hash map as 0
    binmap = [0]*CONST_MAX
    for i in range(0,arr_size):
        temp = sum-arr[i]
        if (temp>=0 and binmap[temp]==1):
            print "Pair with the given sum is", arr[i], "and", temp
# driver program to check the above function
A = [1,4,45,6,10,-8]
n = 16
printPairs(A, len(A), n)

Time Complexity: O(n)

Pair with given sum 16 is (10, 6)

Auxiliary Space: O(R) where R is range of integers.

If range of numbers include negative numbers then also it works. All we have to do for negative numbers is to make everything positive by adding the absolute value of smallest negative integer to all numbers.

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