Given an array of n elements and an integer K, the task is to find the subarray with minimum value of ||a[i] + a[i + 1] + ……. a[j]| – K|. In other words, find the contiguous sub-array whose sum of elements shows minimum deviation from K or the subarray whose absolute sum is closest to K.
Example
Input:: a[] = {1, 3, 7, 10}, K = 15
Output: Subarray {7, 10}
The contiguous sub-array [7, 10] shows minimum deviation of 2 from 15.Input: a[] = {1, 2, 3, 4, 5, 6}, K = 6
Output: Subarray {1, 2, 3}
The contiguous sub-array [1, 2, 3] shows minimum deviation of 0 from 6.
A naive approach would be to check if the sum of each contiguous sub-array and its difference from K.
Below is the implementation of the above approach:
C++
// C++ code to find sub-array whose // sum shows the minimum deviation #include <bits/stdc++.h>using namespace std;int* getSubArray(int arr[], int n, int K) { int i = -1; int j = -1; int currSum = 0; // Starting index, ending index, // Deviation int* result = new int[3]{ i, j, abs(K - abs(currSum)) }; // Iterate i and j to get all subarrays for(i = 0; i < n; i++) { currSum = 0; for(j = i; j < n; j++) { currSum += arr[j]; int currDev = abs(K - abs(currSum)); // Found sub-array with less sum if (currDev < result[2]) { result[0] = i; result[1] = j; result[2] = currDev; } // Exactly same sum if (currDev == 0) return result; } } return result; } // Driver code int main(){ int arr[8] = { 15, -3, 5, 2, 7, 6, 34, -6 }; int n = sizeof(arr) / sizeof(arr[0]); int K = 50; // Array to store return values int* ans = getSubArray(arr, n, K); if (ans[0] == -1) { cout << "The empty array shows " << "minimum Deviation"; } else { for(int i = ans[0]; i <= ans[1]; i++) cout << arr[i] << " "; } return 0;}// This code is contributed by divyeshrabadiya07 |
Java
// Java code to find sub-array whose// sum shows the minimum deviationclass GFG{ public static int[] getSubArray(int[] arr, int n,int K){ int i = -1; int j = -1; int currSum = 0; // Starting index, ending index, Deviation int [] result = { i, j, Math.abs(K - Math.abs(currSum)) }; // Iterate i and j to get all subarrays for(i = 0; i < n; i++) { currSum = 0; for(j = i; j < n; j++) { currSum += arr[j]; int currDev = Math.abs(K - Math.abs(currSum)); // Found sub-array with less sum if(currDev < result[2]) { result[0] = i; result[1] = j; result[2] = currDev; } // Exactly same sum if(currDev == 0) return result; } } return result;}// Driver Code public static void main(String[] args){ int[] arr = { 15, -3, 5, 2, 7, 6, 34, -6 }; int n = arr.length; int K = 50; // Array to store return values int[] ans = getSubArray(arr, n, K); if(ans[0] == -1) { System.out.println("The empty array " + "shows minimum Deviation"); } else { for(int i = ans[0]; i <= ans[1]; i++) System.out.print(arr[i] + " "); }}}// This code is contributed by dadimadhav |
Python
# Python Code to find sub-array whose # sum shows the minimum deviationdef getSubArray(arr, n, K): i = -1 j = -1 currSum = 0 # starting index, ending index, Deviation result = [i, j, abs(K-abs(currSum))] # iterate i and j to get all subarrays for i in range(0, n): currSum = 0 for j in range(i, n): currSum += arr[j] currDev = abs(K-abs(currSum)) # found sub-array with less sum if (currDev < result[2]): result = [i, j, currDev] # exactly same sum if (currDev == 0): return result return result # Driver Codedef main(): arr = [15, -3, 5, 2, 7, 6, 34, -6] n = len(arr) K = 50 [i, j, minDev] = getSubArray(arr, n, K) if(i ==-1): print("The empty array shows minimum Deviation") return 0 for i in range(i, j + 1): print arr[i], main() |
C#
// C# code to find sub-array whose// sum shows the minimum deviationusing System;class GFG{ public static int[] getSubArray(int[] arr, int n, int K){ int i = -1; int j = -1; int currSum = 0; // Starting index, ending index, Deviation int [] result = { i, j, Math.Abs(K - Math.Abs(currSum)) }; // Iterate i and j to get all subarrays for(i = 0; i < n; i++) { currSum = 0; for(j = i; j < n; j++) { currSum += arr[j]; int currDev = Math.Abs(K - Math.Abs(currSum)); // Found sub-array with less sum if (currDev < result[2]) { result[0] = i; result[1] = j; result[2] = currDev; } // Exactly same sum if (currDev == 0) return result; } } return result;}// Driver Code public static void Main(string[] args){ int[] arr = { 15, -3, 5, 2, 7, 6, 34, -6 }; int n = arr.Length; int K = 50; // Array to store return values int[] ans = getSubArray(arr, n, K); if (ans[0] == -1) { Console.Write("The empty array " + "shows minimum Deviation"); } else { for(int i = ans[0]; i <= ans[1]; i++) Console.Write(arr[i] + " "); }}}// This code is contributed by rutvik_56 |
Javascript
<script>// Javascript code to find sub-array whose// sum shows the minimum deviationfunction getSubArray(arr, n, K){ let i = -1; let j = -1; let currSum = 0; // Starting index, ending index, Deviation let result = [ i, j, Math.abs(K - Math.abs(currSum)) ]; // Iterate i and j to get all subarrays for(i = 0; i < n; i++) { currSum = 0; for(j = i; j < n; j++) { currSum += arr[j]; let currDev = Math.abs(K - Math.abs(currSum)); // Found sub-array with less sum if(currDev < result[2]) { result[0] = i; result[1] = j; result[2] = currDev; } // Exactly same sum if(currDev == 0) return result; } } return result;}// driver code let arr = [ 15, -3, 5, 2, 7, 6, 34, -6 ]; let n = arr.length; let K = 50; // Array to store return values let ans = getSubArray(arr, n, K); if(ans[0] == -1) { document.write("The empty array " + "shows minimum Deviation"); } else { for(let i = ans[0]; i <= ans[1]; i++) document.write(arr[i] + " "); } </script> |
Output:
-3 5 2 7 6 34
Complexity Analysis:
- Time Complexity: O(N²)
- Auxiliary Space: O(1)
Efficient Approach:
If the array only consists of non-negative integers, use the sliding window technique to improve the calculation time for sum in each iteration. The sliding window technique reduces the complexity by calculating the new sub-array sum using the previous sub-array sum. Increase the right index till the difference (K-sum) is greater than zero. The first sub-array with negative (K-sum) is considered, and the next sub-array is with left index = i+1(where i is the current right index).
Below is the implementation of the above approach:
C++
// C++ code to find non-negative sub-array// whose sum shows minimum deviation. This// works only if all elements in array are// non-negative#include <bits/stdc++.h>using namespace std;struct Pair{ int f, s, t; Pair(int f, int s, int t) { this->f = f; this->s = s; this->t = t; }};// Function to return the index Pair* getSubArray(int *arr, int n, int K){ int currSum = 0; int prevDif = 0; int currDif = 0; Pair *result = new Pair( -1, -1, abs(K - abs(currSum))); Pair *resultTmp = result; int i = 0; int j = 0; while (i<= j && j<n) { // Add Last element tp currSum currSum += arr[j]; // Save Difference of previous Iteration prevDif = currDif; // Calculate new Difference currDif = K - abs(currSum); // When the Sum exceeds K if (currDif <= 0) { if (abs(currDif) < abs(prevDif)) { // Current Difference greater // in magnitude. Store Temporary // Result resultTmp = new Pair(i, j, currDif); } else { // Difference in Previous was lesser // In previous, Right index = j-1 resultTmp = new Pair(i, j - 1, prevDif); // In next iteration, Left Index Increases // but Right Index remains the Same // Update currSum and i Accordingly currSum -= (arr[i] + arr[j]); i += 1; } } // Case to simply increase Right Index else { resultTmp = new Pair(i, j, currDif); j += 1; } if (abs(resultTmp->t) < abs(result->t)) { // Check if lesser deviation found result = resultTmp; } } return result;}// Driver Codeint main(){ int arr[] = { 15, -3, 5, 2, 7, 6, 34, -6 }; int n = sizeof(arr) / sizeof(arr[0]); int K = 50; Pair *tmp = getSubArray(arr, n, K); int i = tmp->f; int j = tmp->s; int minDev = tmp->t; if (i == -1) { cout << "The empty array shows minimum Deviation" << endl; return 0; } for(int k = i + 1; k < j + 1; k++) { cout << arr[k] << " "; } return 0;}// This code is contributed by pratham76 |
Java
// Java code to find non-negative sub-array// whose sum shows minimum deviation. This// works only if all elements in array are// non-negativeimport java.util.*;class GFG{static class Pair{ int f, s, t; Pair(int f, int s, int t) { this.f = f; this.s = s; this.t = t; }};// Function to return the index static Pair getSubArray(int []arr, int n, int K){ int currSum = 0; int prevDif = 0; int currDif = 0; Pair result = new Pair( -1, -1, Math.abs(K - Math.abs(currSum))); Pair resultTmp = result; int i = 0; int j = 0; while (i<= j && j<n) { // Add Last element tp currSum currSum += arr[j]; // Save Difference of previous Iteration prevDif = currDif; // Calculate new Difference currDif = K - Math.abs(currSum); // When the Sum exceeds K if (currDif <= 0) { if (Math.abs(currDif) < Math.abs(prevDif)) { // Current Difference greater // in magnitude. Store Temporary // Result resultTmp = new Pair(i, j, currDif); } else { // Difference in Previous was lesser // In previous, Right index = j-1 resultTmp = new Pair(i, j - 1, prevDif); // In next iteration, Left Index Increases // but Right Index remains the Same // Update currSum and i Accordingly currSum -= (arr[i] + arr[j]); i += 1; } } // Case to simply increase Right Index else { resultTmp = new Pair(i, j, currDif); j += 1; } if (Math.abs(resultTmp.t) < Math.abs(result.t)) { // Check if lesser deviation found result = resultTmp; } } return result;}// Driver Codepublic static void main(String[] args){ int arr[] = { 15, -3, 5, 2, 7, 6, 34, -6 }; int n = arr.length; int K = 50; Pair tmp = getSubArray(arr, n, K); int i = tmp.f; int j = tmp.s; int minDev = tmp.t; if (i == -1) { System.out.print("The empty array shows minimum Deviation" +"\n"); return ; } for(int k = i + 1; k < j + 1; k++) { System.out.print(arr[k]+ " "); } }}// This code is contributed by 29AjayKumar |
Python
# Python Code to find non-negative # sub-array whose sum shows minimum deviation# This works only if all elements# in array are non-negative# function to return the index def getSubArray(arr, n, K): currSum = 0 prevDif = 0 currDif = 0 result = [-1, -1, abs(K-abs(currSum))] resultTmp = result i = 0 j = 0 while(i<= j and j<n): # Add Last element tp currSum currSum += arr[j] # Save Difference of previous Iteration prevDif = currDif # Calculate new Difference currDif = K - abs(currSum) # When the Sum exceeds K if(currDif <= 0): if abs(currDif) < abs(prevDif): # Current Difference greater in magnitude # Store Temporary Result resultTmp = [i, j, currDif] else: # Difference in Previous was lesser # In previous, Right index = j-1 resultTmp = [i, j-1, prevDif] # In next iteration, Left Index Increases # but Right Index remains the Same # Update currSum and i Accordingly currSum -= (arr[i]+arr[j]) i += 1 # Case to simply increase Right Index else: resultTmp = [i, j, currDif] j += 1 if(abs(resultTmp[2]) < abs(result[2])): # Check if lesser deviation found result = resultTmp return result# Driver Codedef main(): arr = [15, -3, 5, 2, 7, 6, 34, -6] n = len(arr) K = 50 [i, j, minDev] = getSubArray(arr, n, K) if(i ==-1): print("The empty array shows minimum Deviation") return 0 for i in range(i, j+1): print arr[i], main() |
C#
// C# code to find non-negative sub-array// whose sum shows minimum deviation. This// works only if all elements in array are// non-negativeusing System;public class GFG{ class Pair { public int f, s, t; public Pair(int f, int s, int t) { this.f = f; this.s = s; this.t = t; } }; // Function to return the index static Pair getSubArray(int []arr, int n, int K) { int currSum = 0; int prevDif = 0; int currDif = 0; Pair result = new Pair( -1, -1, Math.Abs(K - Math.Abs(currSum))); Pair resultTmp = result; int i = 0; int j = 0; while (i <= j && j < n) { // Add Last element tp currSum currSum += arr[j]; // Save Difference of previous Iteration prevDif = currDif; // Calculate new Difference currDif = K - Math.Abs(currSum); // When the Sum exceeds K if (currDif <= 0) { if (Math.Abs(currDif) < Math.Abs(prevDif)) { // Current Difference greater // in magnitude. Store Temporary // Result resultTmp = new Pair(i, j, currDif); } else { // Difference in Previous was lesser // In previous, Right index = j-1 resultTmp = new Pair(i, j - 1, prevDif); // In next iteration, Left Index Increases // but Right Index remains the Same // Update currSum and i Accordingly currSum -= (arr[i] + arr[j]); i += 1; } } // Case to simply increase Right Index else { resultTmp = new Pair(i, j, currDif); j += 1; } if (Math.Abs(resultTmp.t) < Math.Abs(result.t)) { // Check if lesser deviation found result = resultTmp; } } return result; } // Driver Code public static void Main(String[] args) { int []arr = { 15, -3, 5, 2, 7, 6, 34, -6 }; int n = arr.Length; int K = 50; Pair tmp = getSubArray(arr, n, K); int i = tmp.f; int j = tmp.s; int minDev = tmp.t; if (i == -1) { Console.Write("The empty array shows minimum Deviation" +"\n"); return ; } for(int k = i + 1; k < j + 1; k++) { Console.Write(arr[k]+ " "); } }}// This code is contributed by 29AjayKumar |
Javascript
// JavaScript code to find non-negative sub-array// whose sum shows minimum deviation. This// works only if all elements in array are// non-negativeclass Pair { constructor(f, s, t) { this.f = f; this.s = s; this.t = t; }}// Function to return the index function getSubArray(arr, n, K) { let currSum = 0; let prevDif = 0; let currDif = 0; let result = new Pair(-1, -1, Math.abs(K - Math.abs(currSum))); let resultTmp = result; let i = 0; let j = 0; while (i <= j && j < n) { // Add Last element to currSum currSum += arr[j]; // Save Difference of previous Iteration prevDif = currDif; // Calculate new Difference currDif = K - Math.abs(currSum); // When the Sum exceeds K if (currDif <= 0) { if (Math.abs(currDif) < Math.abs(prevDif)) { // Current Difference greater // in magnitude. Store Temporary // Result resultTmp = new Pair(i, j, currDif); } else { // Difference in Previous was lesser // In previous, Right index = j-1 resultTmp = new Pair(i, j - 1, prevDif); // In next iteration, Left Index Increases // but Right Index remains the Same // Update currSum and i Accordingly currSum -= (arr[i] + arr[j]); i += 1; } } // Case to simply increase Right Index else { resultTmp = new Pair(i, j, currDif); j += 1; } if (Math.abs(resultTmp.t) < Math.abs(result.t)) { // Check if lesser deviation found result = resultTmp; } } return result;}// Driver Codelet arr = [ 15, -3, 5, 2, 7, 6, 34, -6 ];let n = arr.length;let K = 50;let tmp = getSubArray(arr, n, K);let i = tmp.f;let j = tmp.s;let minDev = tmp.t;if (i == -1) { console.log("The empty array shows minimum Deviation");}for(let k = i + 1; k < j + 1; k++) { console.log(arr[k] + " ");}//Code is contributed by dhanshriborse |
Output
-3 5 2 7 6 34
Complexity Analysis:
- Time Complexity: O(N)
- Auxiliary Space: O(1)
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