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Queries to find minimum sum of array elements from either end of an array
  • Difficulty Level : Easy
  • Last Updated : 29 Apr, 2021

Given an array arr[] consisting of N distinct integers and an array queries[] consisting of Q queries, the task is for each query is to find queries[i] in the array and calculate the minimum of sum of array elements from the start and end of the array up to queries[i].

Examples:

Input: arr[] = {2, 3, 6, 7, 4, 5, 30}, Q = 2, queries[] = {6, 5}
Output: 11 27
Explanation:
Query 1: Sum from start = 2 + 3 + 6 = 11. Sum from end = 30 + 5 + 4 + 7 + 6 = 52. Therefore, 11 is the required answer.
Query 2: Sum from start = 27. Sum from end = 35. Therefore, 27 is the required answer.

Input: arr[] = {1, 2, -3, 4}, Q = 2, queries[] = {4, 2}
Output: 4 2

Naive Approach: The simplest approach is to traverse the array upro queries[i] for each query and calculate sum from the end as well as from the start of the array. Finally, print the minimum of the sums obtained.



Below is the implementation of the above approach:

C++




// C++ implementation
// of the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to calclate the minimum
// sum from either end of the arrays
// for the given queries
void calculateQuery(int arr[], int N,
                    int query[], int M)
{
    // Traverse the query[] array
    for (int i = 0; i < M; i++) {
        int X = query[i];
 
        // Stores sum from start
        // and end of the array
        int sum_start = 0, sum_end = 0;
 
        // Calculate distance from start
        for (int j = 0; j < N; j++) {
 
            sum_start += arr[j];
            if (arr[j] == X)
                break;
        }
 
        // Calculate distance from end
        for (int j = N - 1; j >= 0; j--) {
 
            sum_end += arr[j];
            if (arr[j] == X)
                break;
        }
 
        cout << min(sum_end, sum_start) << " ";
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 3, 6, 7, 4, 5, 30 };
    int queries[] = { 6, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int M = sizeof(queries)
            / sizeof(queries[0]);
 
    calculateQuery(arr, N, queries, M);
 
    return 0;
}

Java




// Java implementation of the
// above approach
import java.util.*;
import java.lang.*;
 
public class GFG
{
 
  // Function to calclate the minimum
  // sum from either end of the arrays
  // for the given queries
  static void calculateQuery(int arr[], int N,
                             int query[], int M)
  {
 
    // Traverse the query[] array
    for (int i = 0; i < M; i++)
    {
      int X = query[i];
 
      // Stores sum from start
      // and end of the array
      int sum_start = 0, sum_end = 0;
 
      // Calculate distance from start
      for (int j = 0; j < N; j++)
      {
        sum_start += arr[j];
        if (arr[j] == X)
          break;
      }
 
      // Calculate distance from end
      for (int j = N - 1; j >= 0; j--)
      {
        sum_end += arr[j];
        if (arr[j] == X)
          break;
      }
      System.out.print(Math.min(sum_end, sum_start) + " ");
    }
  }
 
  // Driver Code
  public static void main (String[] args)
  {
    int arr[] = { 2, 3, 6, 7, 4, 5, 30 };
    int queries[] = { 6, 5 };
    int N = arr.length;
    int M = queries.length;
    calculateQuery(arr, N, queries, M);
  }
}
 
// This code is contributed by jana_sayantan.

Python3




# Python 3 implementation
# of the above approach
 
# Function to calclate the minimum
# sum from either end of the arrays
# for the given queries
 
 
def calculateQuery(arr, N, query, M):
 
    # Traverse the query[] array
    for i in range(M):
        X = query[i]
 
        # Stores sum from start
        # and end of the array
        sum_start = 0
        sum_end = 0
 
        # Calculate distance from start
        for j in range(N):
 
            sum_start += arr[j]
            if (arr[j] == X):
                break
 
        # Calculate distance from end
        for j in range(N - 1, -1, -1):
 
            sum_end += arr[j]
            if (arr[j] == X):
                break
        print(min(sum_end, sum_start), end=" ")
 
# Driver Code
if __name__ == "__main__":
 
    arr = [2, 3, 6, 7, 4, 5, 30]
    queries = [6, 5]
    N = len(arr)
    M = len(queries)
 
    calculateQuery(arr, N, queries, M)
 
    # This code is contributed by chitranayal.

C#




// C# implementation
// of the above approach
using System;
class GFG {
 
  // Function to calclate the minimum
  // sum from either end of the arrays
  // for the given queries
  static void calculateQuery(int[] arr, int N,
                             int[] query, int M)
  {
     
    // Traverse the query[] array
    for (int i = 0; i < M; i++) {
      int X = query[i];
 
      // Stores sum from start
      // and end of the array
      int sum_start = 0, sum_end = 0;
 
      // Calculate distance from start
      for (int j = 0; j < N; j++) {
 
        sum_start += arr[j];
        if (arr[j] == X)
          break;
      }
 
      // Calculate distance from end
      for (int j = N - 1; j >= 0; j--) {
 
        sum_end += arr[j];
        if (arr[j] == X)
          break;
      }
 
      Console.Write(Math.Min(sum_end, sum_start) + " ");
    }
  }
 
  // Driver code
  static void Main()
  {
    int[] arr = { 2, 3, 6, 7, 4, 5, 30 };
    int[] queries = { 6, 5 };
    int N = arr.Length;
    int M = queries.Length;
 
    calculateQuery(arr, N, queries, M);
  }
}
 
// This code is contributed by divyesh072019.

Javascript




<script>
 
// Javascript implementation of the above approach
 
// Function to calclate the minimum
// sum from either end of the arrays
// for the given queries
function calculateQuery(arr, N, query, M)
{
     
    // Traverse the query[] array
    for(let i = 0; i < M; i++)
    {
        let X = query[i];
         
        // Stores sum from start
        // and end of the array
        let sum_start = 0, sum_end = 0;
         
        // Calculate distance from start
        for(let j = 0; j < N; j++)
        {
            sum_start += arr[j];
             
            if (arr[j] == X)
                break;
        }
         
        // Calculate distance from end
        for(let j = N - 1; j >= 0; j--)
        {
            sum_end += arr[j];
             
            if (arr[j] == X)
                break;
        }
        document.write(Math.min(
            sum_end, sum_start) + " ");
    }
}
 
// Driver code
let arr = [ 2, 3, 6, 7, 4, 5, 30 ];
let queries = [ 6, 5 ];
let N = arr.length;
let M = queries.length;
 
calculateQuery(arr, N, queries, M);
 
// This code is contributed by suresh07
 
</script>
Output: 
11 27

 

Time Complexity: O(Q * N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized by using extra space and the concept of prefix sum and suffix sum and to answer each query in constant computational complexity. The idea is to preprocess prefix and suffix sums for each index. Follow the steps below to solve the problem:

  • Initialize two variables, say prefix and suffix.
  • Initialize an Unordered Map, say mp, to map array elements as keys to pairs in which the first value gives the prefix sum and the second value gives the suffix sum.
  • Traverse the array and keep adding arr[i] to prefix and store it in the map with arr[i] as key and prefix as value.
  • Traverse the array in reverse and keep adding arr[i] to suffix and store in the map with arr[i] as key and suffix as value.
  • Now traverse the array queries[] and for each query queries[i], print the value of the minimum of mp[queries[i]].first and mp[queries[i]].second as the result.

Below is the implementation of the above approach: 

C++




// C++ implementation
// of the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum sum
// for the given queries
void calculateQuery(int arr[], int N,
                    int query[], int M)
{
 
    // Stores prefix and suffix sums
    int prefix = 0, suffix = 0;
 
    // Stores pairs of prefix and suffix sums
    unordered_map<int, pair<int, int> > mp;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Add element to prefix
        prefix += arr[i];
 
        // Store prefix for each element
        mp[arr[i]].first = prefix;
    }
 
    // Traverse the array in reverse
    for (int i = N - 1; i >= 0; i--) {
 
        // Add element to suffix
        suffix += arr[i];
 
        // Storing suffix for each element
        mp[arr[i]].second = suffix;
    }
 
    // Travere the array queries[]
    for (int i = 0; i < M; i++) {
 
        int X = query[i];
 
        // Minimum of suffix
        // and prefix sums
        cout << min(mp[X].first,
                    mp[X].second)
             << " ";
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 3, 6, 7, 4, 5, 30 };
    int queries[] = { 6, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int M = sizeof(queries) / sizeof(queries[0]);
 
    calculateQuery(arr, N, queries, M);
 
    return 0;
}

Java




// Java implementation
// of the above approach
import java.util.*;
 
class GFG
{
  static class pair<E,P>
  {
    E first;
    P second;
    public pair(E first, P second) 
    {
      this.first = first;
      this.second = second;
    }   
  }
 
  // Function to find the minimum sum
  // for the given queries
  @SuppressWarnings({ "unchecked", "rawtypes" })
  static void calculateQuery(int arr[], int N,
                             int query[], int M)
  {
 
    // Stores prefix and suffix sums
    int prefix = 0, suffix = 0;
 
    // Stores pairs of prefix and suffix sums
    HashMap<Integer, pair > mp = new HashMap<>();
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
      // Add element to prefix
      prefix += arr[i];
 
      // Store prefix for each element
      mp.put(arr[i], new pair(prefix,0));
    }
 
    // Traverse the array in reverse
    for (int i = N - 1; i >= 0; i--) {
 
      // Add element to suffix
      suffix += arr[i];
 
      // Storing suffix for each element
      mp.put(arr[i], new pair(mp.get(arr[i]).first,suffix));
    }
 
    // Travere the array queries[]
    for (int i = 0; i < M; i++) {
 
      int X = query[i];
 
      // Minimum of suffix
      // and prefix sums
      System.out.print(Math.min((int)mp.get(X).first,
                                (int)mp.get(X).second)
                       + " ");
    }
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int arr[] = { 2, 3, 6, 7, 4, 5, 30 };
    int queries[] = { 6, 5 };
    int N = arr.length;
    int M = queries.length;
 
    calculateQuery(arr, N, queries, M);
 
  }
}
 
// This code is contributed by 29AjayKumar

C#




// C# implementation
// of the above approach
using System;
using System.Collections.Generic;
class GFG {
 
  // Function to find the minimum sum
  // for the given queries
  static void calculateQuery(int[] arr, int N,
                             int[] query, int M)
  {
 
    // Stores prefix and suffix sums
    int prefix = 0, suffix = 0;
 
    // Stores pairs of prefix and suffix sums
    Dictionary<int, Tuple<int,int>> mp =
      new Dictionary<int, Tuple<int,int>>();
 
    // Traverse the array
    for (int i = 0; i < N; i++)
    {
 
      // Add element to prefix
      prefix += arr[i];
 
      // Store prefix for each element
      mp[arr[i]] = new Tuple<int,int>(prefix, 0);
    }
 
    // Traverse the array in reverse
    for (int i = N - 1; i >= 0; i--)
    {
 
      // Add element to suffix
      suffix += arr[i];
 
      // Storing suffix for each element
      mp[arr[i]] = new Tuple<int,int>(mp[arr[i]].Item1, suffix);
    }
 
    // Travere the array queries[]
    for (int i = 0; i < M; i++)
    {
      int X = query[i];
 
      // Minimum of suffix
      // and prefix sums
      Console.Write(Math.Min(mp[X].Item1, mp[X].Item2) + " ");
    }
  }
 
  // Driver code
  static void Main() {
    int[] arr = { 2, 3, 6, 7, 4, 5, 30 };
    int[] queries = { 6, 5 };
    int N = arr.Length;
    int M = queries.Length;
 
    calculateQuery(arr, N, queries, M);
  }
}
 
// This code is contributed by divyeshrabadiya07.
Output: 
11 27

 

Time Complexity: O(M + N)
Auxiliary Space: O(N)

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