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| 1 | +package codechallenge.easy.array; |
| 2 | + |
| 3 | +import org.junit.Test; |
| 4 | + |
| 5 | +/** |
| 6 | + * Need to check the partition patter where sum of first half of array is equal |
| 7 | + * to sum of second half |
| 8 | + * input array is [3,4,2,2,2,1] |
| 9 | + * output index of parition element 2 |
| 10 | + * |
| 11 | + * * input array is [3,2,2,5,1] |
| 12 | + * * output should be -1 |
| 13 | + */ |
| 14 | +public class IsArrayPartitioned { |
| 15 | + |
| 16 | + // Test Driven Development by Aseem Jain |
| 17 | + @Test |
| 18 | + public void test() { |
| 19 | + int[] a = {3,4,2,2,2,1}; |
| 20 | + assert 2 == solution1(a); |
| 21 | + |
| 22 | + int[] b = {3,2,2,5,1}; |
| 23 | + assert -1 == solution1(b); |
| 24 | + |
| 25 | + assert 2 == solution2(a); |
| 26 | + |
| 27 | + assert -1 == solution2(b); |
| 28 | + |
| 29 | + } |
| 30 | + |
| 31 | + // Basic solution O(n x n) |
| 32 | + private int solution1(int[] a) { |
| 33 | + |
| 34 | + int fh =0; |
| 35 | + |
| 36 | + for (int i = 0; i < a.length; i++) { |
| 37 | + fh += a[i]; |
| 38 | + int sh =0; |
| 39 | + for (int j = i+1; j < a.length; j++) { |
| 40 | + sh += a[j]; |
| 41 | + } |
| 42 | + if(fh == sh){ |
| 43 | + return i+1; |
| 44 | + } |
| 45 | + } |
| 46 | + |
| 47 | + return -1; |
| 48 | + } |
| 49 | + |
| 50 | + // Optimized solution O(n) |
| 51 | + private int solution2(int[] a) { |
| 52 | + |
| 53 | + int sum =0; |
| 54 | + int fh =0; |
| 55 | + |
| 56 | + for (int i = 0; i < a.length; i++) { |
| 57 | + sum += a[i]; |
| 58 | + } |
| 59 | + |
| 60 | + // to get the partition, first half |
| 61 | + for (int i = 0; i < a.length; i++) { |
| 62 | + fh += a[i]; |
| 63 | + |
| 64 | + if(fh == sum/2){ |
| 65 | + return i+1; |
| 66 | + } |
| 67 | + } |
| 68 | + |
| 69 | + return -1; |
| 70 | + } |
| 71 | + |
| 72 | + |
| 73 | +} |
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