Maximum Strength of K Disjoint Subarrays solution and explanation
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The problem
You are given an array of integers nums with length n, and a positive odd integer k.
Select exactly k disjoint subarrays sub1, sub2, ..., subk from nums such that the last element of subi appears before the first element of sub{i+1} for all 1 <= i <= k-1. The goal is to maximize their combined strength.
The strength of the selected subarrays is defined as:
strength = k * sum(sub1)- (k - 1) * sum(sub2) + (k - 2) * sum(sub3) - ... - 2 * sum(sub{k-1}) + sum(subk)
where sum(subi) is the sum of the elements in the i-th subarray.
Return the maximum possible strength that can be obtained from selecting exactly k disjoint subarrays from nums.
Note that the chosen subarrays don't need to cover the entire array.
Example 1:
Input: nums = [1,2,3,-1,2], k = 3
Output: 22
Explanation:
The best possible way to select 3 subarrays is: nums[0..2], nums[3..3], and nums[4..4]. The strength is calculated as follows:
strength = 3 * (1 + 2 + 3) - 2 * (-1) + 2 = 22
Example 2:
Input: nums = [12,-2,-2,-2,-2], k = 5
Output: 64
Explanation:
The only possible way to select 5 disjoint subarrays is: nums[0..0], nums[1..1], nums[2..2], nums[3..3], and nums[4..4]. The strength is calculated as follows:
strength = 5 * 12 - 4 * (-2) + 3 * (-2) - 2 * (-2) + (-2) = 64
Example 3:
Input: nums = [-1,-2,-3], k = 1
Output: -1
Explanation:
The best possible way to select 1 subarray is: nums[0..0]. The strength is -1.
Constraints:
1 <= n <= 104-109 <= nums[i] <= 1091 <= k <= n1 <= n * k <= 106kis odd.
Approach overview
Problem Overview: You are given an integer array and must select k disjoint subarrays that maximize a special strength value. Each chosen subarray contributes a weighted sum based on its order, so the algorithm must decide both where each segment starts and ends and which segments produce the highest total score.
Approach 1: Dynamic Programming with Prefix Sum (O(nk) time, O(nk) space)
This approach models the problem as a staged decision process using dynamic programming. Precompute a prefix sum array so any subarray sum can be calculated in O(1). Define DP states that track the maximum strength after processing the first i elements while forming j subarrays, along with whether the current element continues an active segment or starts a new one. During iteration, update transitions for extending the current subarray or closing it and starting the next weighted segment.
The key insight is that each subarray contributes a coefficient based on its order (for example (k - j + 1) with alternating signs). While iterating through the array, the DP transition multiplies the current element by the correct coefficient and adds it to the best previous state. This avoids enumerating all possible subarray boundaries. The algorithm runs in O(nk) time because every index updates DP states for up to k segments.
Approach 2: Sliding Window with Priority Queue (O(n log n) time, O(n) space)
An alternative view treats the task as selecting the best weighted subarray contributions while enforcing the disjoint constraint. Iterate through the array while maintaining running sums for candidate windows. A sliding window identifies profitable segments, and a priority queue stores candidate contributions ordered by strength.
Whenever a window becomes beneficial for the current segment weight, push its contribution into the heap. The heap structure allows quick retrieval of the highest‑value segments while ensuring previously chosen segments remain disjoint. As you move the window forward, outdated candidates are removed and new ones are inserted. Each push or pop operation costs O(log n), producing overall complexity of O(n log n).
Recommended for interviews: The dynamic programming approach is the one most interviewers expect. It clearly demonstrates control over state transitions, weighted scoring, and efficient use of prefix sums. Discussing the heap‑based strategy shows deeper problem exploration, but the DP solution with O(nk) complexity is typically considered the canonical solution.
Approach 1: Dynamic Programming Approach
This approach involves using a dynamic programming table to keep track of the maximum strength possible for each configuration of selecting j subarrays ending at each index i. This table can be filled in by considering at each position the optimal previous configuration plus forming a new subarray ending at that position.
def max_strength_of_k_disjoint_subarrays(nums, k):
n = len(nums)
dp = [[-float('inf')] * (n + 1) for _ in range(k + 1)]
dp[0][0] = 0
for j in range(1, k + 1, 2):
for i in range(j, n + 1):
sum = 0
for l in range(i, j - 1, -1):
sum += nums[l - 1]
if dp[j - 1][l - 1] != -float('inf'):
dp[j][i] = max(dp[j][i], dp[j - 1][l - 1] + (sum if j == 1 or j == k else (-sum if (j - 1) % 4 == 1 else sum)))
return max(dp[k])
# Example usage:
nums = [1, 2, 3, -1, 2]
k = 3
print(max_strength_of_k_disjoint_subarrays(nums, k))import java.util.*;
public class MaxSubarrayStrength {
public static int maxStrengthOfKDisjointSubarrays(int[] nums, int k) {
int n = nums.length;
int[][] dp = new int[k + 1][n + 1];
for (int[] d : dp) Arrays.fill(d, Integer.MIN_VALUE);
dp[0][0] = 0;
for (int j = 1; j <= k; j += 2) {
for (int i = j; i <= n; i++) {
int sum = 0;
for (int l = i; l >= j; l--) {
sum += nums[l - 1];
if (dp[j - 1][l - 1] != Integer.MIN_VALUE) {
dp[j][i] = Math.max(dp[j][i], dp[j - 1][l - 1] + ((j == 1 || j == k) ? sum : ((j - 1) % 4 == 1 ? -sum : sum)));
}
}
}
}
return Arrays.stream(dp[k]).max().getAsInt();
}
public static void main(String[] args) {
int[] nums = {1, 2, 3, -1, 2};
int k = 3;
System.out.println(maxStrengthOfKDisjointSubarrays(nums, k));
}
}#include <iostream>
#include <vector>
#include <algorithm>
#include <limits.h>
using namespace std;
int maxStrengthOfKDisjointSubarrays(vector<int>& nums, int k) {
int n = nums.size();
vector<vector<int>> dp(k + 1, vector<int>(n + 1, INT_MIN));
dp[0][0] = 0;
for (int j = 1; j <= k; j += 2) {
for (int i = j; i <= n; i++) {
int sum = 0;
for (int l = i; l >= j; l--) {
sum += nums[l - 1];
if (dp[j - 1][l - 1] != INT_MIN) {
dp[j][i] = max(dp[j][i], dp[j - 1][l - 1] + ((j == 1 || j == k) ? sum : ((j - 1) % 4 == 1 ? -sum : sum)));
}
}
}
}
return *max_element(dp[k].begin(), dp[k].end());
}
int main() {
vector<int> nums = {1, 2, 3, -1, 2};
int k = 3;
cout << maxStrengthOfKDisjointSubarrays(nums, k) << endl;
return 0;
}function maxStrengthOfKDisjointSubarrays(nums, k) {
let n = nums.length;
let dp = Array.from({ length: k + 1 }, () => Array(n + 1).fill(-Infinity));
dp[0][0] = 0;
for (let j = 1; j <= k; j += 2) {
for (let i = j; i <= n; i++) {
let sum = 0;
for (let l = i; l >= j; l--) {
sum += nums[l - 1];
if (dp[j - 1][l - 1] !== -Infinity) {
dp[j][i] = Math.max(dp[j][i], dp[j - 1][l - 1] + ((j === 1 || j === k) ? sum : ((j - 1) % 4 === 1 ? -sum : sum)));
}
}
}
}
return Math.max(...dp[k]);
}
// Example usage:
let nums = [1, 2, 3, -1, 2];
let k = 3;
console.log(maxStrengthOfKDisjointSubarrays(nums, k));#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int maxStrengthOfKDisjointSubarrays(int* nums, int numsSize, int k) {
int** dp = (int**)malloc((k + 1) * sizeof(int*));
for (int i = 0; i <= k; ++i) {
dp[i] = (int*)malloc((numsSize + 1) * sizeof(int));
for (int j = 0; j <= numsSize; ++j) {
dp[i][j] = INT_MIN;
}
}
dp[0][0] = 0;
for (int j = 1; j <= k; j += 2) {
for (int i = j; i <= numsSize; ++i) {
int sum = 0;
for (int l = i; l >= j; --l) {
sum += nums[l - 1];
if (dp[j - 1][l - 1] != INT_MIN) {
dp[j][i] = (dp[j][i] < dp[j - 1][l - 1] + ((j == 1 || j == k) ? sum : ((j - 1) % 4 == 1 ? -sum : sum))) ? dp[j - 1][l - 1] + ((j == 1 || j == k) ? sum : ((j - 1) % 4 == 1 ? -sum : sum)) : dp[j][i];
}
}
}
}
int max_strength = INT_MIN;
for (int i = k; i <= numsSize; ++i) {
max_strength = (max_strength < dp[k][i]) ? dp[k][i] : max_strength;
}
for (int i = 0; i <= k; ++i) {
free(dp[i]);
}
free(dp);
return max_strength;
}
int main() {
int nums[] = {1, 2, 3, -1, 2};
int k = 3;
int numsSize = sizeof(nums) / sizeof(nums[0]);
printf("%d\n", maxStrengthOfKDisjointSubarrays(nums, numsSize, k));
return 0;
}using System;
public class MaxSubarrayStrength {
public static int MaxStrengthOfKDisjointSubarrays(int[] nums, int k) {
int n = nums.Length;
int[,] dp = new int[k + 1, n + 1];
for (int i = 0; i <= k; ++i)
for (int j = 0; j <= n; ++j)
dp[i, j] = int.MinValue;
dp[0, 0] = 0;
for (int j = 1; j <= k; j += 2) {
for (int i = j; i <= n; i++) {
int sum = 0;
for (int l = i; l >= j; l--) {
sum += nums[l - 1];
if (dp[j - 1, l - 1] != int.MinValue) {
dp[j, i] = Math.Max(dp[j, i], dp[j - 1, l - 1] + ((j == 1 || j == k) ? sum : ((j - 1) % 4 == 1 ? -sum : sum)));
}
}
}
}
int max_strength = int.MinValue;
for (int i = k; i <= n; ++i) {
max_strength = Math.Max(max_strength, dp[k, i]);
}
return max_strength;
}
public static void Main() {
int[] nums = {1, 2, 3, -1, 2};
int k = 3;
Console.WriteLine(MaxStrengthOfKDisjointSubarrays(nums, k));
}
}The Python solution uses a list of lists for the DP table, initialized to negative infinity for unattainable strengths, and zero for the base case. The solution iterates over possible subarray ending indices, computes ongoing sums, and updates the DP table by selectively maximizing based on position sign-flipping conditions.
Time complexity: O(k * n^2), Space complexity: O(k * n) where n is the size of nums.
Approach 2: Sliding Window and Priority Queue Approach
This approach leverages a sliding window technique alongside a priority queue (or heap) to efficiently calculate subarray sums and manage the subarray selection such that the alternating strength function is maximized.
import heapq
def max_strength_of_k_disjoint_subarrays(nums, k):
n = len(nums)
prefix_sum = [0] * (n + 1)
for i in range(1, n + 1):
prefix_sum[i] = prefix_sum[i - 1] + nums[i - 1]
max_heap = []
for i in range(n):
for j in range(i + 1, n + 1):
subarray_sum = prefix_sum[j] - prefix_sum[i]
heapq.heappush(max_heap, -subarray_sum)
max_strength = 0
for i in range(k):
if max_heap:
max_strength += (k - i) * -heapq.heappop(max_heap)
return max_strength
# Example usage:
nums = [1, 2, 3, -1, 2]
k = 3
print(max_strength_of_k_disjoint_subarrays(nums, k))import java.util.*;
public class MaxSubarrayStrength {
public static int maxStrengthOfKDisjointSubarrays(int[] nums, int k) {
PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());
int[] prefixSum = new int[nums.length + 1];
for (int i = 1; i <= nums.length; ++i) {
prefixSum[i] = prefixSum[i - 1] + nums[i - 1];
}
for (int i = 0; i < nums.length; ++i) {
for (int j = i + 1; j <= nums.length; ++j) {
int subarray_sum = prefixSum[j] - prefixSum[i];
maxHeap.add(subarray_sum);
}
}
int max_strength = 0;
for (int i = 0; i < k && !maxHeap.isEmpty(); ++i) {
max_strength += (k - i) * maxHeap.poll();
}
return max_strength;
}
public static void main(String[] args) {
int[] nums = {1, 2, 3, -1, 2};
int k = 3;
System.out.println(maxStrengthOfKDisjointSubarrays(nums, k));
}
}#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
int maxStrengthOfKDisjointSubarrays(vector<int>& nums, int k) {
priority_queue<int> maxHeap;
vector<int> prefixSum(nums.size() + 1, 0);
for (int i = 1; i <= nums.size(); ++i) {
prefixSum[i] = prefixSum[i - 1] + nums[i - 1];
}
for (int i = 0; i < nums.size(); ++i) {
for (int j = i + 1; j <= nums.size(); ++j) {
int subarray_sum = prefixSum[j] - prefixSum[i];
maxHeap.push(subarray_sum);
}
}
int max_strength = 0;
for (int i = 0; i < k && !maxHeap.empty(); ++i) {
max_strength += (k - i) * maxHeap.top();
maxHeap.pop();
}
return max_strength;
}
int main() {
vector<int> nums = {1, 2, 3, -1, 2};
int k = 3;
cout << maxStrengthOfKDisjointSubarrays(nums, k) << endl;
return 0;
}function maxStrengthOfKDisjointSubarrays(nums, k) {
let n = nums.length;
let prefixSum = new Array(n + 1).fill(0);
for (let i = 1; i <= n; ++i) {
prefixSum[i] = prefixSum[i - 1] + nums[i - 1];
}
let maxHeap = [];
for (let i = 0; i < n; ++i) {
for (let j = i + 1; j <= n; ++j) {
let subarray_sum = prefixSum[j] - prefixSum[i];
maxHeap.push(subarray_sum);
}
}
maxHeap.sort((a, b) => b - a);
let max_strength = 0;
for (let i = 0; i < k; ++i) {
if (maxHeap.length) {
max_strength += (k - i) * maxHeap.shift();
}
}
return max_strength;
}
// Example usage:
let nums = [1, 2, 3, -1, 2];
let k = 3;
console.log(maxStrengthOfKDisjointSubarrays(nums, k));#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
typedef struct {
int *data;
int capacity;
int size;
} MaxHeap;
MaxHeap* createHeap(int capacity) {
MaxHeap *heap = (MaxHeap*)malloc(sizeof(MaxHeap));
heap->data = (int*)malloc(sizeof(int) * capacity);
heap->capacity = capacity;
heap->size = 0;
return heap;
}
void swap(int *a, int *b) {
int temp = *a;
*a = *b;
*b = temp;
}
void heapifyUp(MaxHeap *heap, int index) {
if (index && heap->data[(index - 1) / 2] < heap->data[index]) {
swap(&heap->data[index], &heap->data[(index - 1) / 2]);
heapifyUp(heap, (index - 1) / 2);
}
}
void heapifyDown(MaxHeap *heap, int index) {
int largest = index;
int left = 2 * index + 1;
int right = 2 * index + 2;
if (left < heap->size && heap->data[left] > heap->data[largest])
largest = left;
if (right < heap->size && heap->data[right] > heap->data[largest])
largest = right;
if (largest != index) {
swap(&heap->data[index], &heap->data[largest]);
heapifyDown(heap, largest);
}
}
void push(MaxHeap *heap, int value) {
if (heap->size < heap->capacity) {
heap->data[heap->size] = value;
heap->size++;
heapifyUp(heap, heap->size - 1);
}
}
int pop(MaxHeap *heap) {
if (heap->size) {
int root = heap->data[0];
heap->data[0] = heap->data[--heap->size];
heapifyDown(heap, 0);
return root;
}
return INT_MIN;
}
void freeHeap(MaxHeap *heap) {
free(heap->data);
free(heap);
}
int maxStrengthOfKDisjointSubarrays(int* nums, int numsSize, int k) {
MaxHeap *heap = createHeap(numsSize);
int *prefixSum = (int*)malloc((numsSize + 1) * sizeof(int));
prefixSum[0] = 0;
for (int i = 1; i <= numsSize; ++i) {
prefixSum[i] = prefixSum[i - 1] + nums[i - 1];
}
int max_strength = INT_MIN;
for (int i = 0; i < numsSize; ++i) {
for (int j = i + 1; j <= numsSize; ++j) {
int subarray_sum = prefixSum[j] - prefixSum[i];
push(heap, subarray_sum);
if (heap->size > k) pop(heap);
}
}
int factor = k;
while (k-- && heap->size) {
max_strength += factor-- * pop(heap);
}
free(heap);
free(prefixSum);
return max_strength;
}
int main() {
int nums[] = {1, 2, 3, -1, 2};
int k = 3;
int numsSize = sizeof(nums) / sizeof(nums[0]);
printf("%d\n", maxStrengthOfKDisjointSubarrays(nums, numsSize, k));
return 0;
}using System;
using System.Collections.Generic;
public class MaxSubarrayStrength {
public static int MaxStrengthOfKDisjointSubarrays(int[] nums, int k) {
List<int> maxHeap = new List<int>();
int[] prefixSum = new int[nums.Length + 1];
for (int i = 1; i <= nums.Length; ++i) {
prefixSum[i] = prefixSum[i - 1] + nums[i - 1];
}
for (int i = 0; i < nums.Length; ++i) {
for (int j = i + 1; j <= nums.Length; ++j) {
int subarray_sum = prefixSum[j] - prefixSum[i];
maxHeap.Add(subarray_sum);
}
}
maxHeap.Sort((a, b) => b.CompareTo(a));
int max_strength = 0;
for (int i = 0; i < k && i < maxHeap.Count; ++i) {
max_strength += (k - i) * maxHeap[i];
}
return max_strength;
}
public static void Main() {
int[] nums = {1, 2, 3, -1, 2};
int k = 3;
Console.WriteLine(MaxStrengthOfKDisjointSubarrays(nums, k));
}
}This Python code applies a negative sign to the heap to simulate a max-heap for obtaining the highest subarray sums. It efficiently manages subarray sum computation through prefix sums and orderly handling via heapq functions.
Time complexity: O(n^2 * log k), Space complexity: O(n) used for prefix sums.
Approach 3: Dynamic Programming
For the ith number nums[i - 1], if it is selected and is in the jth subarray, then its contribution to the answer is nums[i - 1] times (k - j + 1) times (-1)^{j+1}. We denote (-1)^{j+1} as sign, so its contribution to the answer is sign times nums[i - 1] times (k - j + 1).
We define f[i][j][0] as the maximum energy value when selecting j subarrays from the first i numbers, and the ith number is not selected. We define f[i][j][1] as the maximum energy value when selecting j subarrays from the first i numbers, and the ith number is selected. Initially, f[0][0][1] = 0, and the rest of the values are -infty.
When i > 0, we consider how f[i][j] transitions.
If the ith number is not selected, then the i-1th number can either be selected or not selected, so f[i][j][0] = max(f[i-1][j][0], f[i-1][j][1]).
If the ith number is selected, if the i-1th number and the ith number are in the same subarray, then f[i][j][1] = max(f[i][j][1], f[i-1][j][1] + sign times nums[i-1] times (k - j + 1)), otherwise f[i][j][1] = max(f[i][j][1], max(f[i-1][j-1][0], f[i-1][j-1][1]) + sign times nums[i-1] times (k - j + 1)).
The final answer is max(f[n][k][0], f[n][k][1]).
The time complexity is O(n times k), and the space complexity is O(n times k). Where n is the length of the array nums.
function maximumStrength(nums: number[], k: number): number {
const n: number = nums.length;
const f: number[][][] = Array.from({ length: n + 1 }, () =>
Array.from({ length: k + 1 }, () => [-Infinity, -Infinity]),
);
f[0][0][0] = 0;
for (let i = 1; i <= n; i++) {
const x: number = nums[i - 1];
for (let j = 0; j <= k; j++) {
const sign: number = (j & 1) === 1 ? 1 : -1;
const val: number = sign * x * (k - j + 1);
f[i][j][0] = Math.max(f[i - 1][j][0], f[i - 1][j][1]);
f[i][j][1] = Math.max(f[i][j][1], f[i - 1][j][1] + val);
if (j > 0) {
f[i][j][1] = Math.max(f[i][j][1], Math.max(...f[i - 1][j - 1]) + val);
}
}
}
return Math.max(...f[n][k]);
}func maximumStrength(nums []int, k int) int64 {
n := len(nums)
f := make([][][]int64, n+1)
const inf int64 = math.MinInt64 / 2
for i := range f {
f[i] = make([][]int64, k+1)
for j := range f[i] {
f[i][j] = []int64{inf, inf}
}
}
f[0][0][0] = 0
for i := 1; i <= n; i++ {
x := nums[i-1]
for j := 0; j <= k; j++ {
sign := int64(-1)
if j&1 == 1 {
sign = 1
}
val := sign * int64(x) * int64(k-j+1)
f[i][j][0] = max(f[i-1][j][0], f[i-1][j][1])
f[i][j][1] = max(f[i][j][1], f[i-1][j][1]+val)
if j > 0 {
t := max(f[i-1][j-1][0], f[i-1][j-1][1]) + val
f[i][j][1] = max(f[i][j][1], t)
}
}
}
return max(f[n][k][0], f[n][k][1])
}Complexity comparison
| Approach | Complexity |
|---|---|
| Dynamic Programming Approach | Time complexity: |
| Sliding Window and Priority Queue Approach | Time complexity: |
| Dynamic Programming | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Dynamic Programming with Prefix Sum | O(nk) | O(nk) | General case. Best choice in interviews when selecting k disjoint segments with weighted contributions. |
| Sliding Window + Priority Queue | O(n log n) | O(n) | Useful when modeling segments as candidate windows and extracting top contributions dynamically. |
Frequently asked questions
Is Maximum Strength of K Disjoint Subarrays easy or hard?
Maximum Strength of K Disjoint Subarrays is classified as a Hard problem. The challenge comes from combining weighted scoring, multiple disjoint segments, and efficient transitions using dynamic programming and prefix sums.
Maximum Strength of K Disjoint Subarrays Python/Java solution
Most implementations follow the same DP structure across languages. Maintain a DP table indexed by array position and number of segments, compute prefix sums, and update transitions for extending or starting a subarray. The algorithm translates cleanly into Python, Java, C++, C#, and JavaScript.
How to solve Maximum Strength of K Disjoint Subarrays in O(n)?
A strict O(n) solution generally isn't feasible because the algorithm must evaluate transitions for up to k subarrays. The optimal widely used solution runs in O(nk) using dynamic programming with prefix sums. Each step evaluates whether to extend or start a subarray while applying the correct weight.
What is the best approach for Maximum Strength of K Disjoint Subarrays?
Dynamic programming with prefix sums is the most reliable approach. The DP state tracks how many subarrays have been formed and whether the current element extends a segment or starts a new one. Prefix sums allow constant‑time subarray calculations, resulting in an overall time complexity of O(nk).
Is Maximum Strength of K Disjoint Subarrays asked at Google/Amazon/Meta?
Hard dynamic programming problems involving multiple disjoint subarrays appear frequently in interviews at companies like Google, Amazon, and Meta. Variants that maximize k segments or weighted subarray sums test DP state design and prefix sum optimization.
What data structure is used in Maximum Strength of K Disjoint Subarrays?
The main solution relies on arrays for dynamic programming states and prefix sums for constant‑time range sum queries. Some alternative implementations also use a priority queue (heap) to track candidate subarray contributions while maintaining disjoint constraints.
What is the time complexity of Maximum Strength of K Disjoint Subarrays?
The standard dynamic programming solution runs in O(nk) time, where n is the array length and k is the number of required subarrays. Each element updates DP states for all segment counts up to k. Space complexity is typically O(nk) or optimized to O(k) with rolling arrays.