Maximum element present in the array after performing queries to add K to range of indices [L, R]

Last Updated : 20 Jan, 2023

Given an array arr[] consisting of N integers, ( initially set to 0 ) and an array Q[], consisting of queries of the form {l, r, k}, the task for each query is to add K to the indices l to r(both inclusive). After performing all queries, return the maximum element present the array.

Example:

Input: N=10, q[] = {{1, 5, 3}, {4, 8, 7}, {6, 9, 1}}
Output: 10
Explanation:
Initially the array is ? [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
Query1 {1, 5, 3} results in [3, 3, 3, 3, 3, 0, 0, 0, 0, 0]
Query2 {4, 8, 7} results in [3, 3, 3, 10, 10, 7, 7, 7, 0, 0]
Query2 {6, 9, 1} results in [3, 3, 3, 10, 10, 8, 8, 8, 1, 0]
Maximum value in the updated array = 10

Approach: Follow the steps below to solve the problem.

Traverse over the vector of queries and for each query {l, r, k}
- Add k to a[l] and subtract k from a[r+1]
Initialize variable x = 0 to store the running sum and m = INT_MIN to store the maximum value
Traverse the array, add elements to x, and update m.
Print the maximum value m

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;
// Function to find the max sum
// after processing q queries
int max_sum(int a[],
            vector<pair<pair<int, int>, int> > v,
            int q, int n)
{
    // Store the cumulative sum
    int x = 0;
    // Store the maximum sum
    int m = INT_MIN;

    // Iterate over the range 0 to q
    for (int i = 0; i < q; i++) {

        // Variables to extract
        // values from vector
        int p, q, k;

        p = v[i].first.first;
        q = v[i].first.second;
        k = v[i].second;
        a[p] += k;

        if (q + 1 <= n)

            a[q + 1] -= k;
    }

    // Iterate over the range [1, n]
    for (int i = 1; i <= n; i++)

    {
        // Calculate cumulative sum
        x += a[i];

        // Calculate maximum sum
        m = max(m, x);
    }
    // Return the maximum sum after q queries
    return m;
}

// Driver code
int main()
{

    // Stores the size of array
    // and number of queries
    int n = 10, q = 3;

    // Stores the sum
    int a[n + 5] = { 0 };

    // Storing input queries
    vector<pair<pair<int, int>, int> > v(q);
    v[0].first.first = 1;
    v[0].first.second = 5;
    v[0].second = 3;
    v[1].first.first = 4;
    v[1].first.second = 8;
    v[1].second = 7;
    v[2].first.first = 6;
    v[2].first.second = 9;
    v[2].second = 1;

    // Function call to find the maximum sum
    cout << max_sum(a, v, q, n);
    return 0;
}

Java

// Java program for the above approach
import java.lang.*;
import java.util.*;

class GFG{
    
// Function to find the max sum
// after processing q queries
static int max_sum(int a[],
                   ArrayList<ArrayList<Integer>> v,
                   int q, int n)
{
    
    // Store the cumulative sum
    int x = 0;
    
    // Store the maximum sum
    int m = Integer.MIN_VALUE;

    // Iterate over the range 0 to q
    for(int i = 0; i < q; i++)
    {
        
        // Variables to extract
        // values from vector
        int p, qq, k;

        p = v.get(i).get(0);
        qq = v.get(i).get(1);
        k = v.get(i).get(2);
        a[p] += k;

        if (qq + 1 <= n)
            a[qq + 1] -= k;
    }

    // Iterate over the range [1, n]
    for(int i = 1; i <= n; i++)
    {
        
        // Calculate cumulative sum
        x += a[i];

        // Calculate maximum sum
        m = Math.max(m, x);
    }
    
    // Return the maximum sum after q queries
    return m;
}

// Driver code
public static void main(String[] args)
{ 
    
    // Stores the size of array
    // and number of queries
    int n = 10, q = 3;
    
    // Stores the sum
    int[] a = new int[n + 5];
    
    // Storing input queries
    ArrayList<ArrayList<Integer>> v= new ArrayList<>();
    
    for(int i = 0; i < q; i++)
        v.add(new ArrayList<>());
    
    v.get(0).add(1);
    v.get(0).add(5);
    v.get(0).add(3);
    v.get(1).add(4);
    v.get(1).add(8);
    v.get(1).add(7);
    v.get(2).add(6);
    v.get(2).add(9);
    v.get(2).add(1);
    
    // Function call to find the maximum sum
    System.out.println(max_sum(a, v, q, n));
}
}

// This code is contributed by offbeat

Python3

# Python program for the above approach

# Function to find the max sum
# after processing q queries
def max_sum(a, v, q, n):
  
    # Store the cumulative sum
    x = 0;
    
    # Store the maximum sum
    m = -10**9;

    # Iterate over the range 0 to q
    for i in range(q):

        # Variables to extract
        # values from vector
        p = v[i][0][0];
        q = v[i][0][1];
        k = v[i][1];
        a[p] += k;

        if (q + 1 <= n):
            a[q + 1] -= k;

    # Iterate over the range [1, n]
    for i in range(1, n + 1):
        # Calculate cumulative sum
        x += a[i];

        # Calculate maximum sum
        m = max(m, x);

    # Return the maximum sum after q queries
    return m;

# Driver code

# Stores the size of array
# and number of queries
n = 10
q = 3;

# Stores the sum
a = [0] * (n + 5);

# Storing input queries
v = [[[0 for i in range(2)] for x in range(2)] for z in range(q)]
v[0][0][0] = 1;
v[0][0][1] = 5;
v[0][1] = 3;
v[1][0][0] = 4;
v[1][0][1] = 8;
v[1][1] = 7;
v[2][0][0] = 6;
v[2][0][1] = 9;
v[2][1] = 1;

# Function call to find the maximum sum
print(max_sum(a, v, q, n));

# This code is contributed by _saurabh_jaiswal

using System;
using System.Collections.Generic;

class GFG
{
    // Function to find the max sum
    // after processing q queries
    static int max_sum(int[] a,
                       List<List<int>> v,
                       int q, int n)
    {
        // Store the cumulative sum
        int x = 0;

        // Store the maximum sum
        int m = int.MinValue;

        // Iterate over the range 0 to q
        for (int i = 0; i < q; i++)
        {
            // Variables to extract
            // values from vector
            int p, qq, k;

            p = v[i][0];
            qq = v[i][1];
            k = v[i][2];
            a[p] += k;

            if (qq + 1 <= n)
                a[qq + 1] -= k;
        }

        // Iterate over the range [1, n]
        for (int i = 1; i <= n; i++)
        {
            // Calculate cumulative sum
            x += a[i];

            // Calculate maximum sum
            m = Math.Max(m, x);
        }

        // Return the maximum sum after q queries
        return m;
    }

    // Driver code
    public static void Main(string[] args)
    {
        // Stores the size of array
        // and number of queries
        int n = 10, q = 3;

        // Stores the sum
        int[] a = new int[n + 5];

        // Storing input queries
        List<List<int>> v = new List<List<int>>();

        for (int i = 0; i < q; i++)
            v.Add(new List<int>());

        v[0].Add(1);
        v[0].Add(5);
        v[0].Add(3);
        v[1].Add(4);
        v[1].Add(8);
        v[1].Add(7);
        v[2].Add(6);
        v[2].Add(9);
        v[2].Add(1);

        // Function call to find the maximum sum
        Console.WriteLine(max_sum(a, v, q, n));
    }
}

JavaScript

<script>
// Javascript program for the above approach


// Function to find the max sum
// after processing q queries
function max_sum(a, v, q, n) {
    // Store the cumulative sum
    let x = 0;
    // Store the maximum sum
    let m = Number.MIN_SAFE_INTEGER;

    // Iterate over the range 0 to q
    for (let i = 0; i < q; i++) {

        // Variables to extract
        // values from vector
        let p, q, k;

        p = v[i][0][0];
        q = v[i][0][1];
        k = v[i][1];
        a[p] += k;

        if (q + 1 <= n)

            a[q + 1] -= k;
    }

    // Iterate over the range [1, n]
    for (let i = 1; i <= n; i++) {
        // Calculate cumulative sum
        x += a[i];

        // Calculate maximum sum
        m = Math.max(m, x);
    }
    // Return the maximum sum after q queries
    return m;
}

// Driver code


// Stores the size of array
// and number of queries
let n = 10, q = 3;

// Stores the sum
let a = new Array(n + 5).fill(0);

// Storing input queries
let v = new Array(q).fill(0).map(() => new Array(2).fill(0).map(() => new Array(2).fill(0)));
v[0][0][0] = 1;
v[0][0][1] = 5;
v[0][1] = 3;
v[1][0][0] = 4;
v[1][0][1] = 8;
v[1][1] = 7;
v[2][0][0] = 6;
v[2][0][1] = 9;
v[2][1] = 1;

// Function call to find the maximum sum
document.write(max_sum(a, v, q, n));

// This code is contributed by gfgking.
</script>

Output:

Time Complexity: O(N+K) where N is the size of array and K is a number of queries
Space Complexity: O(1)

Perform K of Q queries to maximize the sum of the array elements

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Maximum element present in the array after performing queries to add K to range of indices [L, R]

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