Given an Array[] of N elements and a number K. ( 1 <= K <= N ) . Split the given array into K subarrays (they must cover all the elements). The maximum subarray sum achievable out of K subarrays formed, must be the minimum possible. Find that possible subarray sum.
Examples:
Input : Array[] = {1, 2, 3, 4}, K = 3
Output : 4
Optimal Split is {1, 2}, {3}, {4} . Maximum sum of all subarrays is 4, which is minimum possible for 3 splits.
Input : Array[] = {1, 1, 2} K = 2
Output : 2
Naive approach:
- Idea is to find out all the possibilities.
- To find out all possibilities we use BACKTRACKING.
- At each step, we divide the array into sub-array and find the sum of the sub-array and update the maximum sum
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h> using namespace std;
int ans = 100000000;
// the answer is stored in ans // we call this function solve void solve(int a[], int n, int k, int index, int sum,
int maxsum)
{ // K=1 is the base Case
if (k == 1) {
maxsum = max(maxsum, sum);
sum = 0;
for (int i = index; i < n; i++) {
sum += a[i];
}
// we update maxsum
maxsum = max(maxsum, sum);
// the answer is stored in ans
ans = min(ans, maxsum);
return;
}
sum = 0;
// using for loop to divide the array into K-subarray
for (int i = index; i < n; i++) {
sum += a[i];
// for each subarray we calculate sum ans update
// maxsum
maxsum = max(maxsum, sum);
// calling function again
solve(a, n, k - 1, i + 1, sum, maxsum);
}
} // Driver Code int main()
{ int arr[] = { 1, 2, 3, 4 };
int k = 3; // K divisions
int n = 4; // Size of Array
solve(arr, n, k, 0, 0, 0);
cout << ans << "\n";
} |
C
#include <stdio.h> int ans = 100000000;
// the answer is stored in ans // we call this function solve // max function is used to find max of two elements int max(int a, int b) { return a > b ? a : b; }
// min function is used to find min of two elements int min(int a, int b) { return a < b ? a : b; }
void solve(int a[], int n, int k, int index, int sum,
int maxsum)
{ // K=1 is the base Case
if (k == 1) {
maxsum = max(maxsum, sum);
sum = 0;
for (int i = index; i < n; i++) {
sum += a[i];
}
// we update maxsum
maxsum = max(maxsum, sum);
// the answer is stored in ans
ans = min(ans, maxsum);
return;
}
sum = 0;
// using for loop to divide the array into K-subarray
for (int i = index; i < n; i++) {
sum += a[i];
// for each subarray we calculate sum ans update
// maxsum
maxsum = max(maxsum, sum);
// calling function again
solve(a, n, k - 1, i + 1, sum, maxsum);
}
} // Driver Code int main()
{ int arr[] = { 1, 2, 3, 4 };
int k = 3; // K divisions
int n = 4; // Size of Array
solve(arr, n, k, 0, 0, 0);
printf("%d", ans);
} |
Java
class GFG {
public static int ans = 10000000;
public static void solve(int a[], int n, int k,
int index, int sum, int maxsum)
{
// K=1 is the base Case
if (k == 1) {
maxsum = Math.max(maxsum, sum);
sum = 0;
for (int i = index; i < n; i++) {
sum += a[i];
}
// we update maxsum
maxsum = Math.max(maxsum, sum);
// the answer is stored in ans
ans = Math.min(ans, maxsum);
return;
}
sum = 0;
// using for loop to divide the array into
// K-subarray
for (int i = index; i < n; i++) {
sum += a[i];
// for each subarray we calculate sum ans update
// maxsum
maxsum = Math.max(maxsum, sum);
// calling function again
solve(a, n, k - 1, i + 1, sum, maxsum);
}
}
public static void main(String[] args)
{
int arr[] = { 1, 2, 3, 4 };
int k = 3; // K divisions
int n = 4; // Size of Array
solve(arr, n, k, 0, 0, 0);
System.out.println(ans + "\n");
}
} |
Python3
ans = 10000000
def solve(a, n, k, index, sum, maxsum):
global ans
# K=1 is the base Case
if (k == 1):
maxsum = max(maxsum, sum)
sum = 0
for i in range(index,n):
sum += a[i]
# we update maxsum
maxsum = max(maxsum, sum)
# the answer is stored in ans
ans = min(ans, maxsum)
return
sum = 0
# using for loop to divide the array into
# K-subarray
for i in range(index, n):
sum += a[i]
# for each subarray we calculate sum ans update
# maxsum
maxsum = max(maxsum, sum)
# calling function again
solve(a, n, k - 1, i + 1, sum, maxsum)
# driver code arr = [ 1, 2, 3, 4 ]
k = 3 # K divisions
n = 4 # Size of Array
solve(arr, n, k, 0, 0, 0)
print(ans)
# this code is contributed by shinjanpatra |
C#
using System;
class GFG {
public static int ans = 10000000;
public static void solve(int []a, int n, int k,
int index, int sum, int maxsum)
{
// K=1 is the base Case
if (k == 1) {
maxsum = Math.Max(maxsum, sum);
sum = 0;
for (int i = index; i < n; i++) {
sum += a[i];
}
// we update maxsum
maxsum = Math.Max(maxsum, sum);
// the answer is stored in ans
ans = Math.Min(ans, maxsum);
return;
}
sum = 0;
// using for loop to divide the array into
// K-subarray
for (int i = index; i < n; i++) {
sum += a[i];
// for each subarray we calculate sum ans update
// maxsum
maxsum = Math.Max(maxsum, sum);
// calling function again
solve(a, n, k - 1, i + 1, sum, maxsum);
}
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 1, 2, 3, 4 };
int k = 3; // K divisions
int n = 4; // Size of Array
solve(arr, n, k, 0, 0, 0);
Console.Write(ans + "\n");
}
} // This code is contributed by shivanisinghss2110 |
Javascript
<script> var ans = 10000000;
function solve(a, n, k, index, sum, maxsum)
{
// K=1 is the base Case
if (k == 1) {
maxsum = Math.max(maxsum, sum);
sum = 0;
for (var i = index; i < n; i++) {
sum += a[i];
}
// we update maxsum
maxsum = Math.max(maxsum, sum);
// the answer is stored in ans
ans = Math.min(ans, maxsum);
return;
}
sum = 0;
// using for loop to divide the array into
// K-subarray
for (var i = index; i < n; i++) {
sum += a[i];
// for each subarray we calculate sum ans update
// maxsum
maxsum = Math.max(maxsum, sum);
// calling function again
solve(a, n, k - 1, i + 1, sum, maxsum);
}
}
var arr = [ 1, 2, 3, 4 ];
var k = 3; // K divisions
var n = 4; // Size of Array
solve(arr, n, k, 0, 0, 0);
document.write(ans + "\n");
// this code is contributed by shivanisinghss2110
</script> |
Output
4
Time Complexity: O((N−1)c(K−1) (NOTE: ‘c’ here depicts combinations i.e. ((n-1)!/((n-k)!*(k-1)!)
Where N is the number of elements of the array and K is the number of divisions.
Space Complexity: O(K)
Efficient Approach :
- Idea is to use Binary Search to find an optimal solution.
- For binary search minimum sum can be 1 and the maximum sum can be the sum of all the elements.
- To check if mid is the maximum subarray sum possible. Maintain a count of sub-arrays, include all possible elements in subarray until their sum is less than mid. After this evaluation, if the count is less than or equal to K, then mid is achievable else not. (Since if the count is less than K, we can further divide any subarray its sum will never increase mid ).
- Find the minimum possible value of mid which satisfies the condition.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std;
// Function to check if mid can // be maximum sub - arrays sum bool check(int mid, int array[], int n, int K)
{ int count = 0;
int sum = 0;
for (int i = 0; i < n; i++) {
// If individual element is greater
// maximum possible sum
if (array[i] > mid)
return false;
// Increase sum of current sub - array
sum += array[i];
// If the sum is greater than
// mid increase count
if (sum > mid) {
count++;
sum = array[i];
}
}
count++;
// Check condition
if (count <= K)
return true;
return false;
} // Function to find maximum subarray sum // which is minimum int solve(int array[], int n, int K)
{ int* max = max_element(array, array + n);
int start = *max; //Max subarray sum, considering subarray of length 1
int end = 0;
for (int i = 0; i < n; i++) {
end += array[i]; //Max subarray sum, considering subarray of length n
}
// Answer stores possible
// maximum sub array sum
int answer = 0;
while (start <= end) {
int mid = (start + end) / 2;
// If mid is possible solution
// Put answer = mid;
if (check(mid, array, n, K)) {
answer = mid;
end = mid - 1;
}
else {
start = mid + 1;
}
}
return answer;
} // Driver Code int main()
{ int array[] = { 1, 2, 3, 4 };
int n = sizeof(array) / sizeof(array[0]);
int K = 3;
cout << solve(array, n, K);
} |
Java
// Java implementation of the above approach class GFG {
// Function to check if mid can
// be maximum sub - arrays sum
static boolean check(int mid, int array[], int n, int K)
{
int count = 0;
int sum = 0;
for (int i = 0; i < n; i++) {
// If individual element is greater
// maximum possible sum
if (array[i] > mid)
return false;
// Increase sum of current sub - array
sum += array[i];
// If the sum is greater than
// mid increase count
if (sum > mid) {
count++;
sum = array[i];
}
}
count++;
// Check condition
if (count <= K)
return true;
return false;
}
// Function to find maximum subarray sum
// which is minimum
static int solve(int array[], int n, int K)
{
int start = 1;
for (int i = 0; i < n; ++i) {
if (array[i] > start)
start = array[i]; //Max subarray sum, considering subarray of length 1
}
int end = 0;
for (int i = 0; i < n; i++) {
end += array[i]; //Max subarray sum, considering subarray of length n
}
// Answer stores possible
// maximum sub array sum
int answer = 0;
while (start <= end) {
int mid = (start + end) / 2;
// If mid is possible solution
// Put answer = mid;
if (check(mid, array, n, K)) {
answer = mid;
end = mid - 1;
}
else {
start = mid + 1;
}
}
return answer;
}
// Driver Code
public static void main(String[] args)
{
int array[] = { 1, 2, 3, 4 };
int n = array.length;
int K = 3;
System.out.println(solve(array, n, K));
}
} // This code is contributed by AnkitRai01 |
Python3
# Python 3 implementation of the above approach # Function to check if mid can # be maximum sub - arrays sum def check(mid, array, n, K):
count = 0
sum = 0
for i in range(n):
# If individual element is greater
# maximum possible sum
if (array[i] > mid):
return False
# Increase sum of current sub - array
sum += array[i]
# If the sum is greater than
# mid increase count
if (sum > mid):
count += 1
sum = array[i]
count += 1
# Check condition
if (count <= K):
return True
return False
# Function to find maximum subarray sum # which is minimum def solve(array, n, K):
start = max(array) #Max subarray sum, considering subarray of length 1
end = 0
for i in range(n):
end += array[i] #Max subarray sum, considering subarray of length n
# Answer stores possible
# maximum sub array sum
answer = 0
while (start <= end):
mid = (start + end) // 2
# If mid is possible solution
# Put answer = mid;
if (check(mid, array, n, K)):
answer = mid
end = mid - 1
else:
start = mid + 1
return answer
# Driver Code if __name__ == '__main__':
array = [1, 2, 3, 4]
n = len(array)
K = 3
print(solve(array, n, K))
# This code is contributed by # Surendra_Gangwar |
C#
// C# implementation of the above approach using System;
class GFG
{ // Function to check if mid can
// be maximum sub - arrays sum
static Boolean check(int mid, int []array,
int n, int K)
{
int count = 0;
int sum = 0;
for (int i = 0; i < n; i++)
{
// If individual element is greater
// maximum possible sum
if (array[i] > mid)
return false;
// Increase sum of current sub - array
sum += array[i];
// If the sum is greater than
// mid increase count
if (sum > mid)
{
count++;
sum = array[i];
}
}
count++;
// Check condition
if (count <= K)
return true;
return false;
}
// Function to find maximum subarray sum
// which is minimum
static int solve(int []array, int n, int K)
{
int start = 1;
for (int i = 0; i < n; ++i) {
if (array[i] > start)
start = array[i]; //Max subarray sum, considering subarray of length 1
}
int end = 0;
for (int i = 0; i < n; i++)
{
end += array[i]; //Max subarray sum, considering subarray of length n
}
// Answer stores possible
// maximum sub array sum
int answer = 0;
while (start <= end)
{
int mid = (start + end) / 2;
// If mid is possible solution
// Put answer = mid;
if (check(mid, array, n, K))
{
answer = mid;
end = mid - 1;
}
else
{
start = mid + 1;
}
}
return answer;
}
// Driver Code
public static void Main (String[] args)
{
int []array = { 1, 2, 3, 4 };
int n = array.Length ;
int K = 3;
Console.WriteLine(solve(array, n, K));
}
} // This code is contributed by Princi Singh |
Javascript
<script> // Javascript implementation of the above approach // Function to check if mid can // be maximum sub - arrays sum function check(mid, array, n, K)
{ var count = 0;
var sum = 0;
for (var i = 0; i < n; i++) {
// If individual element is greater
// maximum possible sum
if (array[i] > mid)
return false;
// Increase sum of current sub - array
sum += array[i];
// If the sum is greater than
// mid increase count
if (sum > mid) {
count++;
sum = array[i];
}
}
count++;
// Check condition
if (count <= K)
return true;
return false;
} // Function to find maximum subarray sum // which is minimum function solve(array, n, K)
{ var max = array.reduce((a,b)=>Math.max(a,b));
var start = max; //Max subarray sum, considering subarray of length 1
var end = 0;
for (var i = 0; i < n; i++) {
end += array[i]; //Max subarray sum, considering subarray of length n
}
// Answer stores possible
// maximum sub array sum
var answer = 0;
while (start <= end) {
var mid = parseInt((start + end) / 2);
// If mid is possible solution
// Put answer = mid;
if (check(mid, array, n, K)) {
answer = mid;
end = mid - 1;
}
else {
start = mid + 1;
}
}
return answer;
} // Driver Code var array = [1, 2, 3, 4];
var n = array.length;
var K = 3;
document.write( solve(array, n, K)); </script> |
Output
4
Time Complexity : O(N*log(Sum))
Where N is the number of elements of the array and Sum is the sum of all the elements of the array.
Auxiliary Space: O(1) as no extra space has been used.