Maximum value in an array after m range increment operations
Last Updated :
06 Mar, 2025
Consider an array of size n with all initial values as 0. We need to perform the following m range increment operations.
increment(a, b, k) : Increment values from 'a' to 'b' by 'k'.
After m operations, we need to calculate the maximum of the values in the array.
Examples:
Input : n = 5 m = 3
a = 0, b = 1, k = 100
a = 1, b = 4, k = 100
a = 2, b = 3, k = 100
Output : 200
Explanation:
Initially array = {0, 0, 0, 0, 0}
After first operation: {100, 100, 0, 0, 0}
After second operation: {100, 200, 100, 100, 100}
After third operation {100, 200, 200, 200, 100}
Maximum element after m operations is 200.
Input : n = 4 m = 3
a = 1, b = 2, k = 603
a = 0, b = 0, k = 286
a = 3, b = 3, k = 882
Output : 882
Explanation:
Initially array = {0, 0, 0, 0}
After first operation: {0, 603, 603, 0}
After second operation: {286, 603, 603, 0}
After third operation: {286, 603, 603, 882}
Maximum element after m operations is 882.
[Naive Approach] - One by One Increment
A naive method is to perform each operation on the given range and then, at last, find the maximum number.
C++
#include <bits/stdc++.h>
using namespace std;
// Function to find the maximum element after
// m operations
int findMax(int n, vector<int>& a, vector<int>& b,
vector<int>& k) {
vector<int> arr(n, 0);
// start performing m operations
for (int i = 0; i < a.size(); i++) {
// Store lower and upper index i.e. range
int lowerbound = a[i];
int upperbound = b[i];
// Add 'k[i]' value at this operation to
// whole range
for (int j = lowerbound; j <= upperbound; j++)
arr[j] += k[i];
}
// Find maximum value after all operations and
// return
int res = INT_MIN;
for (int i = 0; i < n; i++)
res = max(res, arr[i]);
return res;
}
// Driver code
int main() {
// Number of queries
int n = 5;
vector<int> a = {0, 1, 2};
vector<int> b = {1, 4, 3};
// value of k to be added at each operation
vector<int> k = {100, 100, 100};
cout << "Maximum value after 'm' operations is "
<< findMax(n, a, b, k);
return 0;
}
import java.util.Arrays;
public class Main {
// Function to find the maximum element after
// m operations
public static int findMax(int n, int[] a, int[] b, int[] k) {
int[] arr = new int[n];
// start performing m operations
for (int i = 0; i < a.length; i++) {
// Store lower and upper index i.e. range
int lowerbound = a[i];
int upperbound = b[i];
// Add 'k[i]' value at this operation to
// whole range
for (int j = lowerbound; j <= upperbound; j++)
arr[j] += k[i];
}
// Find maximum value after all operations and
// return
int res = Integer.MIN_VALUE;
for (int i = 0; i < n; i++)
res = Math.max(res, arr[i]);
return res;
}
// Driver code
public static void main(String[] args) {
// Number of queries
int n = 5;
int[] a = {0, 1, 2};
int[] b = {1, 4, 3};
// value of k to be added at each operation
int[] k = {100, 100, 100};
System.out.println("Maximum value after 'm' operations is " + findMax(n, a, b, k));
}
}
# Function to find the maximum element after
# m operations
def find_max(n, a, b, k):
arr = [0] * n
# start performing m operations
for i in range(len(a)):
# Store lower and upper index i.e. range
lowerbound = a[i]
upperbound = b[i]
# Add 'k[i]' value at this operation to
# whole range
for j in range(lowerbound, upperbound + 1):
arr[j] += k[i]
# Find maximum value after all operations and
# return
res = float('-inf')
for i in range(n):
res = max(res, arr[i])
return res
# Driver code
if __name__ == '__main__':
# Number of queries
n = 5
a = [0, 1, 2]
b = [1, 4, 3]
# value of k to be added at each operation
k = [100, 100, 100]
print("Maximum value after 'm' operations is ", find_max(n, a, b, k))
using System;
using System.Linq;
class Program {
// Function to find the maximum element after
// m operations
public static int FindMax(int n, int[] a, int[] b, int[] k) {
int[] arr = new int[n];
// start performing m operations
for (int i = 0; i < a.Length; i++) {
// Store lower and upper index i.e. range
int lowerbound = a[i];
int upperbound = b[i];
// Add 'k[i]' value at this operation to
// whole range
for (int j = lowerbound; j <= upperbound; j++)
arr[j] += k[i];
}
// Find maximum value after all operations and
// return
int res = int.MinValue;
for (int i = 0; i < n; i++)
res = Math.Max(res, arr[i]);
return res;
}
// Driver code
static void Main() {
// Number of queries
int n = 5;
int[] a = {0, 1, 2};
int[] b = {1, 4, 3};
// value of k to be added at each operation
int[] k = {100, 100, 100};
Console.WriteLine("Maximum value after 'm' operations is " + FindMax(n, a, b, k));
}
}
// Function to find the maximum element after
// m operations
function findMax(n, a, b, k) {
let arr = new Array(n).fill(0);
// start performing m operations
for (let i = 0; i < a.length; i++) {
// Store lower and upper index i.e. range
let lowerbound = a[i];
let upperbound = b[i];
// Add 'k[i]' value at this operation to
// whole range
for (let j = lowerbound; j <= upperbound; j++)
arr[j] += k[i];
}
// Find maximum value after all operations and
// return
let res = Number.MIN_SAFE_INTEGER;
for (let i = 0; i < n; i++)
res = Math.max(res, arr[i]);
return res;
}
// Driver code
let n = 5;
let a = [0, 1, 2];
let b = [1, 4, 3];
// value of k to be added at each operation
let k = [100, 100, 100];
console.log("Maximum value after 'm' operations is " + findMax(n, a, b, k));
OutputMaximum value after 'm' operations is 200
Time Complexity: O(m * max(range)). Here max(range) means maximum elements to which k is added in a single operation.
Auxiliary space: O(n)
[Expected Approach] - Using Prefix Sum
The idea is similar to this post.
Perform two things in a single operation:
- Add k-value to the only lower_bound of a range.
- Reduce the upper_bound + 1 index by a k-value.
After all operations, add all values, check the maximum sum, and print the maximum sum.
C++
#include <bits/stdc++.h>
using namespace std;
// Function to find maximum value after 'm' operations
int findMax(int n, vector<int>& a, vector<int>& b, vector<int>& k)
{
vector<int> arr(n + 1, 0);
// Start performing 'm' operations
for (int i = 0; i < a.size(); i++)
{
// Store lower and upper index i.e. range
int lowerbound = a[i];
int upperbound = b[i];
// Add k to the lower_bound
arr[lowerbound] += k[i];
// Reduce upper_bound+1 indexed value by k
if (upperbound + 1 < arr.size())
arr[upperbound + 1] -= k[i];
}
// Find maximum sum possible from all values
int sum = 0, res = INT_MIN;
for (int i = 0; i < n; ++i)
{
sum += arr[i];
res = max(res, sum);
}
return res;
}
// Driver code
int main()
{
// Number of values
int n = 5;
vector<int> a = {0, 1, 2};
vector<int> b = {1, 4, 3};
vector<int> k = {100, 100, 100};
cout << "Maximum value after 'm' operations is "
<< findMax(n, a, b, k);
return 0;
}
// Import necessary packages
import java.util.*;
// Function to find maximum value after 'm' operations
public class Main {
public static int findMax(int n, int[] a, int[] b, int[] k) {
int[] arr = new int[n + 1];
// Start performing 'm' operations
for (int i = 0; i < a.length; i++) {
// Store lower and upper index i.e. range
int lowerbound = a[i];
int upperbound = b[i];
// Add k to the lower_bound
arr[lowerbound] += k[i];
// Reduce upper_bound+1 indexed value by k
if (upperbound + 1 < arr.length)
arr[upperbound + 1] -= k[i];
}
// Find maximum sum possible from all values
int sum = 0, res = Integer.MIN_VALUE;
for (int i = 0; i < n; ++i) {
sum += arr[i];
res = Math.max(res, sum);
}
return res;
}
// Driver code
public static void main(String[] args) {
// Number of values
int n = 5;
int[] a = {0, 1, 2};
int[] b = {1, 4, 3};
int[] k = {100, 100, 100};
System.out.println("Maximum value after 'm' operations is " + findMax(n, a, b, k));
}
}
# Function to find maximum value after 'm' operations
def find_max(n, a, b, k):
arr = [0] * (n + 1)
# Start performing 'm' operations
for i in range(len(a)):
# Store lower and upper index i.e. range
lowerbound = a[i]
upperbound = b[i]
# Add k to the lower_bound
arr[lowerbound] += k[i]
# Reduce upper_bound+1 indexed value by k
if upperbound + 1 < len(arr):
arr[upperbound + 1] -= k[i]
# Find maximum sum possible from all values
sum = 0
res = float('-inf')
for i in range(n):
sum += arr[i]
res = max(res, sum)
return res
# Driver code
if __name__ == '__main__':
# Number of values
n = 5
a = [0, 1, 2]
b = [1, 4, 3]
k = [100, 100, 100]
print("Maximum value after 'm' operations is ", find_max(n, a, b, k))
// Import necessary packages
using System;
using System.Linq;
// Function to find maximum value after 'm' operations
class GfG {
public static int FindMax(int n, int[] a, int[] b, int[] k) {
int[] arr = new int[n + 1];
// Start performing 'm' operations
for (int i = 0; i < a.Length; i++) {
// Store lower and upper index i.e. range
int lowerbound = a[i];
int upperbound = b[i];
// Add k to the lower_bound
arr[lowerbound] += k[i];
// Reduce upper_bound+1 indexed value by k
if (upperbound + 1 < arr.Length)
arr[upperbound + 1] -= k[i];
}
// Find maximum sum possible from all values
int sum = 0, res = int.MinValue;
for (int i = 0; i < n; ++i) {
sum += arr[i];
res = Math.Max(res, sum);
}
return res;
}
// Driver code
public static void Main(string[] args) {
// Number of values
int n = 5;
int[] a = {0, 1, 2};
int[] b = {1, 4, 3};
int[] k = {100, 100, 100};
Console.WriteLine("Maximum value after 'm' operations is " + FindMax(n, a, b, k));
}
}
// Function to find maximum value after 'm' operations
function findMax(n, a, b, k) {
let arr = new Array(n + 1).fill(0);
// Start performing 'm' operations
for (let i = 0; i < a.length; i++) {
// Store lower and upper index i.e. range
let lowerbound = a[i];
let upperbound = b[i];
// Add k to the lower_bound
arr[lowerbound] += k[i];
// Reduce upper_bound+1 indexed value by k
if (upperbound + 1 < arr.length) {
arr[upperbound + 1] -= k[i];
}
}
// Find maximum sum possible from all values
let sum = 0;
let res = Number.NEGATIVE_INFINITY;
for (let i = 0; i < n; i++) {
sum += arr[i];
res = Math.max(res, sum);
}
return res;
}
// Driver code
const n = 5;
const a = [0, 1, 2];
const b = [1, 4, 3];
const k = [100, 100, 100];
console.log("Maximum value after 'm' operations is ", findMax(n, a, b, k));
OutputMaximum value after 'm' operations is 200
Time complexity: O(m + n)
Auxiliary space: O(n)
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