Given an array of n integers, find the sum of f(a[i], a[j]) of all pairs (i, j) such that (1 <= i < j <= n).
f(a[i], a[j]):
If |a[j]-a[i]| > 1 f(a[i], a[j]) = a[j] - a[i] Else // if |a[j]-a[i]| <= 1 f(a[i], a[j]) = 0
Examples:
Input : 6 6 4 4 Output : -8 Explanation: All pairs are: (6 - 6) + (6 - 6) + (6 - 6) + (4 - 6) + (4 - 6) + (4 - 6) + (4 - 6) + (4 - 4) + (4 - 4) = -8 Input: 1 2 3 1 3 Output: 4 Explanation: the pairs that add up are: (3, 1), (3, 1) to give 4, rest all pairs according to condition gives 0.
A naive approach is to iterate through all pairs and calculate f(a[i], a[j]) and summing it while traversing in two nested loops will give us our answer.
Time Complexity: O(n^2)
A efficient approach will be to use a map/hash function to keep a count of every occurring numbers and then traverse through the list. While traversing through the list, we multiply the count of numbers that are before it and the number itself. Then subtract this result with the pre-sum of the number before that number to get the sum of difference of all pairs possible with that number. To remove all pairs whose absolute difference is <=1, simply subtract the count of occurrence of (number-1) and (number+1) from the previously computed sum. Here we subtract count of (number-1) from the computed sum as it had been previously added to the sum, and we add (number+1) count since the negative has been added to the pre-computed sum of all pairs.
Time Complexity : O(n)
Below is the implementation of the above approach :
C++
// CPP program to calculate the // sum of f(a[i], aj]) #include <bits/stdc++.h> using namespace std; // Function to calculate the sum int sum(int a[], int n) { // map to keep a count of occurrences unordered_map<int, int> cnt; // Traverse in the list from start to end // number of times a[i] can be in a pair and // to get the difference we subtract pre_sum. int ans = 0, pre_sum = 0; for (int i = 0; i < n; i++) { ans += (i * a[i]) - pre_sum; pre_sum += a[i]; // if the (a[i]-1) is present then // subtract that value as f(a[i], a[i]-1)=0 if (cnt[a[i] - 1]) ans -= cnt[a[i] - 1]; // if the (a[i]+1) is present then // add that value as f(a[i], a[i]-1)=0 // here we add as a[i]-(a[i]-1)<0 which would // have been added as negative sum, so we add // to remove this pair from the sum value if (cnt[a[i] + 1]) ans += cnt[a[i] + 1]; // keeping a counter for every element cnt[a[i]]++; } return ans; } // Driver code int main() { int a[] = { 1, 2, 3, 1, 3 }; int n = sizeof(a) / sizeof(a[0]); cout << sum(a, n); return 0; } |
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Java
// Java program to calculate // the sum of f(a[i], aj]) import java.util.*; public class GfG { // Function to calculate the sum public static int sum(int a[], int n) { // Map to keep a count of occurrences Map<Integer,Integer> cnt = new HashMap<Integer,Integer>(); // Traverse in the list from start to end // number of times a[i] can be in a pair and // to get the difference we subtract pre_sum int ans = 0, pre_sum = 0; for (int i = 0; i < n; i++) { ans += (i * a[i]) - pre_sum; pre_sum += a[i]; // If the (a[i]-1) is present then subtract // that value as f(a[i], a[i]-1) = 0 if (cnt.containsKey(a[i] - 1)) ans -= cnt.get(a[i] - 1); // If the (a[i]+1) is present then // add that value as f(a[i], a[i]-1)=0 // here we add as a[i]-(a[i]-1)<0 which would // have been added as negative sum, so we add // to remove this pair from the sum value if (cnt.containsKey(a[i] + 1)) ans += cnt.get(a[i] + 1); // keeping a counter for every element if(cnt.containsKey(a[i])) { cnt.put(a[i], cnt.get(a[i]) + 1); } else { cnt.put(a[i], 1); } } return ans; } // Driver code public static void main(String args[]) { int a[] = { 1, 2, 3, 1, 3 }; int n = a.length; System.out.println(sum(a, n)); } } // This code is contributed by Swetank Modi |
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Python3
# Python3 program to calculate the # sum of f(a[i], aj]) # Function to calculate the sum def sum(a, n): # map to keep a count of occurrences cnt = dict() # Traverse in the list from start to end # number of times a[i] can be in a pair and # to get the difference we subtract pre_sum. ans = 0 pre_sum = 0 for i in range(n): ans += (i * a[i]) - pre_sum pre_sum += a[i] # if the (a[i]-1) is present then # subtract that value as f(a[i], a[i]-1)=0 if (a[i] - 1) in cnt: ans -= cnt[a[i] - 1] # if the (a[i]+1) is present then add that # value as f(a[i], a[i]-1)=0 here we add # as a[i]-(a[i]-1)<0 which would have been # added as negative sum, so we add to remove # this pair from the sum value if (a[i] + 1) in cnt: ans += cnt[a[i] + 1] # keeping a counter for every element if a[i] not in cnt: cnt[a[i]] = 0 cnt[a[i]] += 1 return ans # Driver Code if __name__ == '__main__': a = [1, 2, 3, 1, 3] n = len(a) print(sum(a, n)) # This code is contributed by # SHUBHAMSINGH10 |
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C#
using System; using System.Collections.Generic; // C# program to calculate // the sum of f(a[i], aj]) public class GfG { // Function to calculate the sum public static int sum(int[] a, int n) { // Map to keep a count of occurrences IDictionary<int, int> cnt = new Dictionary<int, int>(); // Traverse in the list from start to end // number of times a[i] can be in a pair and // to get the difference we subtract pre_sum int ans = 0, pre_sum = 0; for (int i = 0; i < n; i++) { ans += (i * a[i]) - pre_sum; pre_sum += a[i]; // If the (a[i]-1) is present then subtract // that value as f(a[i], a[i]-1) = 0 if (cnt.ContainsKey(a[i] - 1)) { ans -= cnt[a[i] - 1]; } // If the (a[i]+1) is present then // add that value as f(a[i], a[i]-1)=0 // here we add as a[i]-(a[i]-1)<0 which would // have been added as negative sum, so we add // to remove this pair from the sum value if (cnt.ContainsKey(a[i] + 1)) { ans += cnt[a[i] + 1]; } // keeping a counter for every element if (cnt.ContainsKey(a[i])) { cnt[a[i]] = cnt[a[i]] + 1; } else { cnt[a[i]] = 1; } } return ans; } // Driver code public static void Main(string[] args) { int[] a = new int[] {1, 2, 3, 1, 3}; int n = a.Length; Console.WriteLine(sum(a, n)); } } // This code is contributed by Shrikant13 |
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Output :
4
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