Hello Everyone,
Given a sorted array (sorted in non-decreasing order) of positive numbers, find the smallest positive integer value that cannot be represented as sum of elements of any subset of given set.
Expected time complexity is O(n).
Examples:
Input: arr[] = {1, 3, 6, 10, 11, 15};
Output: 2
Input: arr[] = {1, 1, 1, 1};
Output: 5
Input: arr[] = {1, 1, 3, 4};
Output: 10
Input: arr[] = {1, 2, 5, 10, 20, 40};
Output: 4
Input: arr[] = {1, 2, 3, 4, 5, 6};
Output: 22
A Simple Solution is to start from value 1 and check all values one by one if they can sum to values in the given array. This solution is very inefficient as it reduces to subset sum problem which is a well known NP Complete Problem.
We can solve this problem in O(n) time using a simple loop. Let the input array be arr[0…n-1]. We initialize the result as 1 (smallest possible outcome) and traverse the given array. Let the smallest element that cannot be represented by elements at indexes from 0 to (i-1) be ‘res’, there are following two possibilities when we consider element at index i:
1) We decide that ‘res’ is the final result: If arr[i] is greater than ‘res’, then we found the gap which is ‘res’ because the elements after arr[i] are also going to be greater than ‘res’.
2) The value of ‘res’ is incremented after considering arr[i]: The value of ‘res’ is incremented by arr[i] (why? If elements from 0 to (i-1) can represent 1 to ‘res-1’, then elements from 0 to i can represent from 1 to ‘res + arr[i] – 1’ be adding ‘arr[i]’ to all subsets that represent 1 to ‘res’)
Following is the implementation of above idea.
// C++ program to find the smallest positive value that cannot be
// represented as sum of subsets of a given sorted array
#include <bits/stdc++.h>
using namespace std;
// Returns the smallest number that cannot be represented as sum
// of subset of elements from set represented by sorted array arr[0..n-1]
int findSmallest( int arr[], int n)
{
int res = 1; // Initialize result
// Traverse the array and increment 'res' if arr[i] is
// smaller than or equal to 'res'.
for ( int i = 0; i < n && arr[i] <= res; i++)
res = res + arr[i];
return res;
}
// Driver program to test above function
int main()
{
int arr1[] = {1, 3, 4, 5};
int n1 = sizeof (arr1)/ sizeof (arr1[0]);
cout << findSmallest(arr1, n1) << endl;
int arr2[] = {1, 2, 6, 10, 11, 15};
int n2 = sizeof (arr2)/ sizeof (arr2[0]);
cout << findSmallest(arr2, n2) << endl;
int arr3[] = {1, 1, 1, 1};
int n3 = sizeof (arr3)/ sizeof (arr3[0]);
cout << findSmallest(arr3, n3) << endl;
int arr4[] = {1, 1, 3, 4};
int n4 = sizeof (arr4)/ sizeof (arr4[0]);
cout << findSmallest(arr4, n4) << endl;
return 0;
}
Output:
2 4 5 10
Time Complexity of above program is O(n).