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Commit 0f6c82c

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+ All pairs whose sum is x (unsorted arrays)
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#http://www.geeksforgeeks.org/given-two-unsorted-arrays-find-pairs-whose-sum-x/
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#Hashing
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def sum_pairs(array1: list, array2: list, req_sum: int):
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sum_lists = []
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#hash array1 for later searching
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hash1 = { x:x for x in array1 }
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for y in array2:
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try:
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#if hash1[req_sum-y] exists
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x = hash1[req_sum - y]
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sum_lists.append([x,y])
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except:
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#the other number for pairing with y not found
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continue
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return sum_lists
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#main
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if __name__=="__main__":
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array1 = list(map(int,input().split()))
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array2 = list(map(int,input().split()))
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req_sum=int(input())
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pairs = sum_pairs(array1,array2,req_sum)
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print("Sum pairs for", req_sum,"are")
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print([pair for pair in pairs])
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"""
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Input Explanation :
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- List 1
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- List 2
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- Required sum
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Input :
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-1 -2 4 -6 5 7
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6 3 4 0
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8
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Output :
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Sum pairs for 8 are
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[[5, 3], [4, 4]]
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"""
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'''
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Given two unsorted arrays of distinct elements, the task is to find all pairs from both arrays whose sum is equal to x.
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Examples:
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Input : arr1[] = {-1, -2, 4, -6, 5, 7}
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arr2[] = {6, 3, 4, 0}
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x = 8
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Output : 4 4
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5 3
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Input : arr1[] = {1, 2, 4, 5, 7}
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arr2[] = {5, 6, 3, 4, 8}
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x = 9
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Output : 1 8
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4 5
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5 4
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Asked in : Amazon
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Recommended: Please solve it on "PRACTICE " first, before moving on to the solution.
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A Naive approach is to simply run two loops and pick elements from both arrays. One by one check that both elements sum is equal to given value x or not.
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// C++ program to find all pairs in both arrays
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// whose sum is equal to given value x
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#include<bits/stdc++.h>
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using namespace std;
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// Function to print all pairs in both arrays
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// whose sum is equal to given value x
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void findPairs(int arr1[], int arr2[], int n,
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int m, int x)
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{
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for (int i=0; i<n; i++)
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for (int j=0; j<m; j++)
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if (arr1[i] + arr2[j] == x)
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cout << arr1[i] << " "
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<< arr2[j] << endl;
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}
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// Driver code
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int main()
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{
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int arr1[] = {1, 2, 3, 7, 5, 4};
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int arr2[] = {0, 7, 4, 3, 2, 1};
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int n = sizeof(arr1)/sizeof(int);
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int m = sizeof(arr2)/sizeof(int);
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int x = 8;
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findPairs(arr1, arr2, n, m, x);
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return 0;
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}
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Run on IDE
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Output:
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1 7
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7 1
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5 3
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4 4
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Time Complexity : O(n^2)
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Auxiliary Space : O(1)
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An Efficient solution of this problem is to hashing. Hash table is implemented using unordered_set in C++. We store all first array elements in hash table. For elements of second array, we subtract every element from x and check the result in hash table. If result is present, we print the element and key in hash (which is an element of first array).
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C++Java
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// C++ program to find all pair in both arrays
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// whose sum is equal to given value x
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#include<bits/stdc++.h>
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using namespace std;
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// Function to find all pairs in both arrays
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// whose sum is equal to given value x
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void findPairs(int arr1[], int arr2[], int n,
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int m, int x)
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{
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// Insert all elements of first array in a hash
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unordered_set<int> s;
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for (int i=0; i<n; i++)
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s.insert(arr1[i]);
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// Subtract sum from second array elements one
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// by one and check it's present in array first
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// or not
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for (int j=0; j<m; j++)
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if (s.find(x - arr2[j]) != s.end())
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cout << x-arr2[j] << " "
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<< arr2[j] << endl;
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}
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// Driver code
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int main()
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{
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int arr1[] = {1, 0, -4, 7, 6, 4};
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int arr2[] = {0 ,2, 4, -3, 2, 1};
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int x = 8;
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int n = sizeof(arr1)/sizeof(int);
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int m = sizeof(arr2)/sizeof(int);
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findPairs(arr1, arr2, n, m, x);
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return 0;
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}
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Run on IDE
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Output:
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6 2
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4 4
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6 2
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7 1
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Time Complexity : O(n)
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Auxiliary Space : O(n)
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'''

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