K-th Largest Sum Contiguous Subarray
Given an array of integers. Write a program to find the K-th largest sum of contiguous subarray within the array of numbers that has both negative and positive numbers.
Examples:
Input: a[] = {20, -5, -1}, K = 3
Output: 14
Explanation: All sum of contiguous subarrays are (20, 15, 14, -5, -6, -1)
so the 3rd largest sum is 14.Input: a[] = {10, -10, 20, -40}, k = 6
Output: -10
Explanation: The 6th largest sum among
sum of all contiguous subarrays is -10.
Brute force Approach: Store all the contiguous sums in another array and sort it and print the Kth largest. But in the case of the number of elements being large, the array in which we store the contiguous sums will run out of memory as the number of contiguous subarrays will be large (quadratic order)
Kth largest sum contiguous subarray using Min-Heap:
The key idea is to store the pre-sum of the array in a sum[] array. One can find the sum of contiguous subarray from index i to j as sum[j] – sum[i-1]. Now generate all possible contiguous subarray sums and push them into the Min-Heap only if the size of Min-Heap is less than K or the current sum is greater than the root of the Min-Heap. In the end, the root of the Min-Heap is the required answer
Follow the given steps to solve the problem using the above approach:
- Create a prefix sum array of the input array
- Create a Min-Heap that stores the subarray sum
- Iterate over the given array using the variable i such that 1 <= i <= N, here i denotes the starting point of the subarray
- Create a nested loop inside this loop using a variable j such that i <= j <= N, here j denotes the ending point of the subarray
- Calculate the sum of the current subarray represented by i and j, using the prefix sum array
- If the size of the Min-Heap is less than K, then push this sum into the heap
- Otherwise, if the current sum is greater than the root of the Min-Heap, then pop out the root and push the current sum into the Min-Heap
- Create a nested loop inside this loop using a variable j such that i <= j <= N, here j denotes the ending point of the subarray
- Now the root of the Min-Heap denotes the Kth largest sum, Return it
Below is the implementation of the above approach:
C++
// C++ program to find the K-th largest sum// of subarray#include <bits/stdc++.h>using namespace std;// Function to calculate Kth largest element// in contiguous subarray sumint kthLargestSum(int arr[], int N, int K){ // array to store prefix sums int sum[N + 1]; sum[0] = 0; sum[1] = arr[0]; for (int i = 2; i <= N; i++) sum[i] = sum[i - 1] + arr[i - 1]; // priority_queue of min heap priority_queue<int, vector<int>, greater<int> > Q; // loop to calculate the contiguous subarray // sum position-wise for (int i = 1; i <= N; i++) { // loop to traverse all positions that // form contiguous subarray for (int j = i; j <= N; j++) { // calculates the contiguous subarray // sum from j to i index int x = sum[j] - sum[i - 1]; // if queue has less than k elements, // then simply push it if (Q.size() < K) Q.push(x); else { // it the min heap has equal to // k elements then just check // if the largest kth element is // smaller than x then insert // else its of no use if (Q.top() < x) { Q.pop(); Q.push(x); } } } } // the top element will be then kth // largest element return Q.top();}// Driver's codeint main(){ int a[] = { 10, -10, 20, -40 }; int N = sizeof(a) / sizeof(a[0]); int K = 6; // Function call cout << kthLargestSum(a, N, K); return 0;} |
Java
// Java program to find the K-th// largest sum of subarrayimport java.util.*;class KthLargestSumSubArray { // function to calculate Kth largest // element in contiguous subarray sum static int kthLargestSum(int arr[], int N, int K) { // array to store prefix sums int sum[] = new int[N + 1]; sum[0] = 0; sum[1] = arr[0]; for (int i = 2; i <= N; i++) sum[i] = sum[i - 1] + arr[i - 1]; // priority_queue of min heap PriorityQueue<Integer> Q = new PriorityQueue<Integer>(); // loop to calculate the contiguous subarray // sum position-wise for (int i = 1; i <= N; i++) { // loop to traverse all positions that // form contiguous subarray for (int j = i; j <= N; j++) { // calculates the contiguous subarray // sum from j to i index int x = sum[j] - sum[i - 1]; // if queue has less then k elements, // then simply push it if (Q.size() < K) Q.add(x); else { // it the min heap has equal to // k elements then just check // if the largest kth element is // smaller than x then insert // else its of no use if (Q.peek() < x) { Q.poll(); Q.add(x); } } } } // the top element will be then kth // largest element return Q.poll(); } // Driver's Code public static void main(String[] args) { int a[] = new int[] { 10, -10, 20, -40 }; int N = a.length; int K = 6; // Function call System.out.println(kthLargestSum(a, N, K)); }}/* This code is contributed by Danish Kaleem */ |
Python3
# Python program to find the K-th largest sum# of subarrayimport heapq# function to calculate Kth largest element# in contiguous subarray sumdef kthLargestSum(arr, N, K): # array to store prefix sums sum = [] sum.append(0) sum.append(arr[0]) for i in range(2, N + 1): sum.append(sum[i - 1] + arr[i - 1]) # priority_queue of min heap Q = [] heapq.heapify(Q) # loop to calculate the contiguous subarray # sum position-wise for i in range(1, N + 1): # loop to traverse all positions that # form contiguous subarray for j in range(i, N + 1): x = sum[j] - sum[i - 1] # if queue has less then k elements, # then simply push it if len(Q) < K: heapq.heappush(Q, x) else: # it the min heap has equal to # k elements then just check # if the largest kth element is # smaller than x then insert # else its of no use if Q[0] < x: heapq.heappop(Q) heapq.heappush(Q, x) # the top element will be then kth # largest element return Q[0]# Driver's codeif __name__ == "__main__": a = [10, -10, 20, -40] N = len(a) K = 6 # Function call print(kthLargestSum(a, N, K))# This code is contributed by Kumar Suman |
C#
// C# program to find the K-th// largest sum of subarrayusing System;using System.Collections.Generic;public class KthLargestSumSubArray { // function to calculate Kth largest // element in contiguous subarray sum static int kthLargestSum(int[] arr, int N, int K) { // array to store prefix sums int[] sum = new int[N + 1]; sum[0] = 0; sum[1] = arr[0]; for (int i = 2; i <= N; i++) sum[i] = sum[i - 1] + arr[i - 1]; // priority_queue of min heap List<int> Q = new List<int>(); // loop to calculate the contiguous subarray // sum position-wise for (int i = 1; i <= N; i++) { // loop to traverse all positions that // form contiguous subarray for (int j = i; j <= N; j++) { // calculates the contiguous subarray // sum from j to i index int x = sum[j] - sum[i - 1]; // if queue has less then k elements, // then simply push it if (Q.Count < K) Q.Add(x); else { // it the min heap has equal to // k elements then just check // if the largest kth element is // smaller than x then insert // else its of no use Q.Sort(); if (Q[0] < x) { Q.RemoveAt(0); Q.Add(x); } } Q.Sort(); } } // the top element will be then Kth // largest element return Q[0]; } // Driver's Code public static void Main(String[] args) { int[] a = new int[] { 10, -10, 20, -40 }; int N = a.Length; int K = 6; // Function call Console.WriteLine(kthLargestSum(a, N, K)); }}// This code contributed by Rajput-Ji |
JavaScript
// Javascript program to find the k-th largest sum// of subarray// function to calculate kth largest element// in contiguous subarray sumfunction kthLargestSum(arr, n, k){ // array to store prefix sums var sum = new Array(n + 1); sum[0] = 0; sum[1] = arr[0]; for (var i = 2; i <= n; i++) sum[i] = sum[i - 1] + arr[i - 1]; // priority_queue of min heap var Q = []; // loop to calculate the contiguous subarray // sum position-wise for (var i = 1; i <= n; i++) { // loop to traverse all positions that // form contiguous subarray for (var j = i; j <= n; j++) { // calculates the contiguous subarray // sum from j to i index var x = sum[j] - sum[i - 1]; // if queue has less then k elements, // then simply push it if (Q.length < k) Q.push(x); else { // it the min heap has equal to // k elements then just check // if the largest kth element is // smaller than x then insert // else its of no use Q.sort(); if (Q[0] < x) { Q.pop(); Q.push(x); } } Q.sort(); } } // the top element will be then kth // largest element return Q[0];}// Driver program to test above functionvar a = [ 10, -10, 20, -40 ];var n = a.length;var k = 6;// calls the function to find out the// k-th largest sumdocument.write(kthLargestSum(a, n, k)); |
C++
// C++ program to find the K-th largest sum// of subarray#include <bits/stdc++.h>using namespace std;// Function to calculate Kth largest element// in contiguous subarray sumint kthLargestSum(int arr[], int N, int K){ // array to store prefix sums int sum[N + 1]; sum[0] = 0; sum[1] = arr[0]; for (int i = 2; i <= N; i++) sum[i] = sum[i - 1] + arr[i - 1]; // priority_queue of min heap priority_queue<int, vector<int>, greater<int> > Q; // loop to calculate the contiguous subarray // sum position-wise for (int i = 1; i <= N; i++) { // loop to traverse all positions that // form contiguous subarray for (int j = i; j <= N; j++) { // calculates the contiguous subarray // sum from j to i index int x = sum[j] - sum[i - 1]; // if queue has less than k elements, // then simply push it if (Q.size() < K) Q.push(x); else { // it the min heap has equal to // k elements then just check // if the largest kth element is // smaller than x then insert // else its of no use if (Q.top() < x) { Q.pop(); Q.push(x); } } } } // the top element will be then kth // largest element return Q.top();}// Driver's codeint main(){ int a[] = { 10, -10, 20, -40 }; int N = sizeof(a) / sizeof(a[0]); int K = 6; // Function call cout << kthLargestSum(a, N, K); return 0;} |
Java
// Java program to find the K-th// largest sum of subarrayimport java.util.*;class KthLargestSumSubArray { // function to calculate Kth largest // element in contiguous subarray sum static int kthLargestSum(int arr[], int N, int K) { // array to store prefix sums int sum[] = new int[N + 1]; sum[0] = 0; sum[1] = arr[0]; for (int i = 2; i <= N; i++) sum[i] = sum[i - 1] + arr[i - 1]; // priority_queue of min heap PriorityQueue<Integer> Q = new PriorityQueue<Integer>(); // loop to calculate the contiguous subarray // sum position-wise for (int i = 1; i <= N; i++) { // loop to traverse all positions that // form contiguous subarray for (int j = i; j <= N; j++) { // calculates the contiguous subarray // sum from j to i index int x = sum[j] - sum[i - 1]; // if queue has less than k elements, // then simply push it if (Q.size() < K) Q.add(x); else { // it the min heap has equal to // k elements then just check // if the largest kth element is // smaller than x then insert // else its of no use if (Q.peek() < x) { Q.poll(); Q.add(x); } } } } // the top element will be then kth // largest element return Q.poll(); } // Driver's Code public static void main(String[] args) { int a[] = new int[] { 10, -10, 20, -40 }; int N = a.length; int K = 6; // Function call System.out.println(kthLargestSum(a, N, K)); }}/* This code is contributed by Danish Kaleem */ |
Python3
# Python program to find the K-th largest sum# of subarrayimport heapq# function to calculate Kth largest element# in contiguous subarray sumdef kthLargestSum(arr, N, K): # array to store prefix sums sum = [] sum.append(0) sum.append(arr[0]) for i in range(2, N + 1): sum.append(sum[i - 1] + arr[i - 1]) # priority_queue of min heap Q = [] heapq.heapify(Q) # loop to calculate the contiguous subarray # sum position-wise for i in range(1, N + 1): # loop to traverse all positions that # form contiguous subarray for j in range(i, N + 1): x = sum[j] - sum[i - 1] # if queue has less than k elements, # then simply push it if len(Q) < K: heapq.heappush(Q, x) else: # it the min heap has equal to # k elements then just check # if the largest kth element is # smaller than x then insert # else its of no use if Q[0] < x: heapq.heappop(Q) heapq.heappush(Q, x) # the top element will be then kth # largest element return Q[0]# Driver's codeif __name__ == "__main__": a = [10, -10, 20, -40] N = len(a) K = 6 # Function call print(kthLargestSum(a, N, K))# This code is contributed by Kumar Suman |
C#
// C# program to find the K-th// largest sum of subarrayusing System;using System.Collections.Generic;public class KthLargestSumSubArray { // function to calculate Kth largest // element in contiguous subarray sum static int kthLargestSum(int[] arr, int N, int K) { // array to store prefix sums int[] sum = new int[N + 1]; sum[0] = 0; sum[1] = arr[0]; for (int i = 2; i <= N; i++) sum[i] = sum[i - 1] + arr[i - 1]; // priority_queue of min heap List<int> Q = new List<int>(); // loop to calculate the contiguous subarray // sum position-wise for (int i = 1; i <= N; i++) { // loop to traverse all positions that // form contiguous subarray for (int j = i; j <= N; j++) { // calculates the contiguous subarray // sum from j to i index int x = sum[j] - sum[i - 1]; // if queue has less than k elements, // then simply push it if (Q.Count < K) Q.Add(x); else { // it the min heap has equal to // k elements then just check // if the largest kth element is // smaller than x then insert // else its of no use Q.Sort(); if (Q[0] < x) { Q.RemoveAt(0); Q.Add(x); } } Q.Sort(); } } // the top element will be then Kth // largest element return Q[0]; } // Driver's Code public static void Main(String[] args) { int[] a = new int[] { 10, -10, 20, -40 }; int N = a.Length; int K = 6; // Function call Console.WriteLine(kthLargestSum(a, N, K)); }}// This code contributed by Rajput-Ji |
JavaScript
// Javascript program to find the k-th largest sum// of subarray// function to calculate kth largest element// in contiguous subarray sumfunction kthLargestSum(arr, n, k){ // array to store prefix sums var sum = new Array(n + 1); sum[0] = 0; sum[1] = arr[0]; for (var i = 2; i <= n; i++) sum[i] = sum[i - 1] + arr[i - 1]; // priority_queue of min heap var Q = []; // loop to calculate the contiguous subarray // sum position-wise for (var i = 1; i <= n; i++) { // loop to traverse all positions that // form contiguous subarray for (var j = i; j <= n; j++) { // calculates the contiguous subarray // sum from j to i index var x = sum[j] - sum[i - 1]; // if queue has less than k elements, // then simply push it if (Q.length < k) Q.push(x); else { // it the min heap has equal to // k elements then just check // if the largest kth element is // smaller than x then insert // else its of no use Q.sort(); if (Q[0] < x) { Q.pop(); Q.push(x); } } Q.sort(); } } // the top element will be then kth // largest element return Q[0];}// Driver program to test above functionvar a = [ 10, -10, 20, -40 ];var n = a.length;var k = 6;// calls the function to find out the// k-th largest sumdocument.write(kthLargestSum(a, n, k)); |
Output
-10
Time Complexity: O(N2 log K)
Auxiliary Space: O(N), but this can be reduced to O(K) for min-heap and we can store the prefix sum array in the input array itself as it is of no use.
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