Queries to find minimum sum of array elements from either end of an array
Last Updated :
01 Sep, 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 upto 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++
#include <bits/stdc++.h>
using namespace std;
void calculateQuery( int arr[], int N,
int query[], int M)
{
for ( int i = 0; i < M; i++) {
int X = query[i];
int sum_start = 0, sum_end = 0;
for ( int j = 0; j < N; j++) {
sum_start += arr[j];
if (arr[j] == X)
break ;
}
for ( int j = N - 1; j >= 0; j--) {
sum_end += arr[j];
if (arr[j] == X)
break ;
}
cout << min(sum_end, sum_start) << " " ;
}
}
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
import java.util.*;
import java.lang.*;
public class GFG
{
static void calculateQuery( int arr[], int N,
int query[], int M)
{
for ( int i = 0 ; i < M; i++)
{
int X = query[i];
int sum_start = 0 , sum_end = 0 ;
for ( int j = 0 ; j < N; j++)
{
sum_start += arr[j];
if (arr[j] == X)
break ;
}
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) + " " );
}
}
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);
}
}
|
Python3
def calculateQuery(arr, N, query, M):
for i in range (M):
X = query[i]
sum_start = 0
sum_end = 0
for j in range (N):
sum_start + = arr[j]
if (arr[j] = = X):
break
for j in range (N - 1 , - 1 , - 1 ):
sum_end + = arr[j]
if (arr[j] = = X):
break
print ( min (sum_end, sum_start), end = " " )
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)
|
C#
using System;
class GFG {
static void calculateQuery( int [] arr, int N,
int [] query, int M)
{
for ( int i = 0; i < M; i++) {
int X = query[i];
int sum_start = 0, sum_end = 0;
for ( int j = 0; j < N; j++) {
sum_start += arr[j];
if (arr[j] == X)
break ;
}
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) + " " );
}
}
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);
}
}
|
Javascript
<script>
function calculateQuery(arr, N, query, M)
{
for (let i = 0; i < M; i++)
{
let X = query[i];
let sum_start = 0, sum_end = 0;
for (let j = 0; j < N; j++)
{
sum_start += arr[j];
if (arr[j] == X)
break ;
}
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) + " " );
}
}
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);
</script>
|
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++
#include <bits/stdc++.h>
using namespace std;
void calculateQuery( int arr[], int N,
int query[], int M)
{
int prefix = 0, suffix = 0;
unordered_map< int , pair< int , int > > mp;
for ( int i = 0; i < N; i++) {
prefix += arr[i];
mp[arr[i]].first = prefix;
}
for ( int i = N - 1; i >= 0; i--) {
suffix += arr[i];
mp[arr[i]].second = suffix;
}
for ( int i = 0; i < M; i++) {
int X = query[i];
cout << min(mp[X].first,
mp[X].second)
<< " " ;
}
}
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
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;
}
}
@SuppressWarnings ({ "unchecked" , "rawtypes" })
static void calculateQuery( int arr[], int N,
int query[], int M)
{
int prefix = 0 , suffix = 0 ;
HashMap<Integer, pair > mp = new HashMap<>();
for ( int i = 0 ; i < N; i++) {
prefix += arr[i];
mp.put(arr[i], new pair(prefix, 0 ));
}
for ( int i = N - 1 ; i >= 0 ; i--) {
suffix += arr[i];
mp.put(arr[i], new pair(mp.get(arr[i]).first,suffix));
}
for ( int i = 0 ; i < M; i++) {
int X = query[i];
System.out.print(Math.min(( int )mp.get(X).first,
( int )mp.get(X).second)
+ " " );
}
}
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);
}
}
|
Python3
def calculateQuery(arr, N, query, M):
prefix = 0
suffix = 0
mp = {}
for i in range (N):
prefix + = arr[i]
mp[arr[i]] = [prefix, 0 ]
for i in range (N - 1 , - 1 , - 1 ):
suffix + = arr[i]
mp[arr[i]] = [mp[arr[i]][ 0 ], suffix]
for i in range (M):
X = query[i]
print ( min (mp[X][ 0 ], mp[X][ 1 ]), end = " " )
arr = [ 2 , 3 , 6 , 7 , 4 , 5 , 30 ]
queries = [ 6 , 5 ]
N = len (arr)
M = len (queries)
calculateQuery(arr, N, queries, M)
|
C#
using System;
using System.Collections.Generic;
class GFG {
static void calculateQuery( int [] arr, int N,
int [] query, int M)
{
int prefix = 0, suffix = 0;
Dictionary< int , Tuple< int , int >> mp =
new Dictionary< int , Tuple< int , int >>();
for ( int i = 0; i < N; i++)
{
prefix += arr[i];
mp[arr[i]] = new Tuple< int , int >(prefix, 0);
}
for ( int i = N - 1; i >= 0; i--)
{
suffix += arr[i];
mp[arr[i]] = new Tuple< int , int >(mp[arr[i]].Item1, suffix);
}
for ( int i = 0; i < M; i++)
{
int X = query[i];
Console.Write(Math.Min(mp[X].Item1, mp[X].Item2) + " " );
}
}
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);
}
}
|
Javascript
<script>
function calculateQuery(arr, N, query, M)
{
var prefix = 0, suffix = 0;
var mp = new Map();
for ( var i = 0; i < N; i++)
{
prefix += arr[i];
if (!mp.has(arr[i]))
{
mp.set(arr[i], [0, 0]);
}
var tmp = mp.get(arr[i]);
tmp[0] = prefix;
mp.set(arr[i], tmp);
}
for ( var i = N - 1; i >= 0; i--)
{
suffix += arr[i];
if (!mp.has(arr[i]))
{
mp.set(arr[i], [0, 0]);
}
var tmp = mp.get(arr[i]);
tmp[1] = suffix;
mp.set(arr[i], tmp);
}
for ( var i = 0; i < M; i++)
{
var X = query[i];
var tmp = mp.get(X);
document.write(Math.min(tmp[0], tmp[1]) + " " );
}
}
var arr = [ 2, 3, 6, 7, 4, 5, 30 ];
var queries = [ 6, 5 ];
var N = arr.length;
var M = queries.length;
calculateQuery(arr, N, queries, M);
</script>
|
Time Complexity: O(M + N)
Auxiliary Space: O(N)
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