57 lines
1.8 KiB
Java
57 lines
1.8 KiB
Java
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/*This is a Java Program to find maximum subarray sum of an array.
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A subarray is a continuous portion of an array. The following program uses binary search approach to find the maximum subarray sum.
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The time complexity of the following program is O (n log n).*/
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/*
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* Java Program to Find the maximum subarray sum using Binary Search approach
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*/
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import java.util.Scanner;
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public class MaxSubarraySum2
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{
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public static void main(String[] args)
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{
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Scanner scan = new Scanner(System.in);
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System.out.println("Enter number of elements in array");
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int N = scan.nextInt();
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int[] arr = new int[ N ];
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/* Accept N elements */
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System.out.println("Enter "+ N +" elements");
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for (int i = 0; i < N; i++)
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arr[i] = scan.nextInt();
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System.out.println("Max sub array sum = "+ max_sum(arr));
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}
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public static int max_sum(int[] arr)
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{
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return max_sum(arr, 0, arr.length - 1);
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}
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public static int max_sum(int[] arr, int low, int mid, int high)
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{
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int l = Integer.MIN_VALUE, r = Integer.MIN_VALUE;
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for (int i = mid, sum = 0; i >= low; i--)
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if ((sum += arr[i]) > l)
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l = sum;
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for (int i = mid +1, sum = 0; i <= high; i++)
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if ((sum += arr[i]) > r)
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r = sum;
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return l + r;
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}
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public static int max_sum(int[] arr, int low, int high)
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{
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if (low == high)
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return arr[low];
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int mid = (low + high)/2;
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int max1 = max_sum(arr, low, mid);
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int max2 = max_sum(arr, mid + 1, high);
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int max3 = max_sum(arr, low, mid, high);
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return Math.max(Math.max(max1, max2), max3);
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}
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}
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/*
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Enter number of elements in array
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8
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Enter 8 elements
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20 5 3 -2 -1 -5 43 24
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Max sub array sum = 87
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