package com.fishercoder.solutions;

import java.util.HashSet;
import java.util.Set;

/**
 * 548. Split Array with Equal Sum
 *
 * Given an array with n integers, you need to find if there are triplets (i, j, k) which satisfies following conditions:
 * 0 < i, i + 1 < j, j + 1 < k < n - 1
 * Sum of subarrays (0, i - 1), (i + 1, j - 1), (j + 1, k - 1) and (k + 1, n - 1) should be equal.
 * where we define that subarray (L, R) represents a slice of the original array starting from the element indexed L to the element indexed R.

 Example:
 Input: [1,2,1,2,1,2,1]
 Output: True
 Explanation:
 i = 1, j = 3, k = 5.
 sum(0, i - 1) = sum(0, 0) = 1
 sum(i + 1, j - 1) = sum(2, 2) = 1
 sum(j + 1, k - 1) = sum(4, 4) = 1
 sum(k + 1, n - 1) = sum(6, 6) = 1

 Note:
 1 <= n <= 2000.
 Elements in the given array will be in range [-1,000,000, 1,000,000].
 */
public class _548 {

    public boolean splitArray(int[] nums) {
        int len = nums.length;
        if (len < 7) {
            return false;
        }
        int[] sum = new int[len];
        sum[0] = nums[0];
        for (int i = 1; i < len; i++) {
            sum[i] = sum[i - 1] + nums[i];
        }
        for (int j = 3; j < len - 3; j++) {
            Set<Integer> set = new HashSet<>();
            for (int i = 1; i < j - 1; i++) {
                if (sum[i - 1] == sum[j - 1] - sum[i]) {
                    set.add(sum[i - 1]);
                }
            }
            for (int k = j + 2; k < len - 1; k++) {
                if (sum[len - 1] - sum[k] == sum[k - 1] - sum[j] && set.contains(sum[k - 1] - sum[j])) {
                    return true;
                }
            }
        }
        return false;
    }

}

package com.fishercoder.solutions;

import java.util.HashSet;
import java.util.Set;

/**
 * 548. Split Array with Equal Sum
 *
 * Given an array with n integers, you need to find if there are triplets (i, j, k) which satisfies following conditions:
 * 0 < i, i + 1 < j, j + 1 < k < n - 1
 * Sum of subarrays (0, i - 1), (i + 1, j - 1), (j + 1, k - 1) and (k + 1, n - 1) should be equal.
 * where we define that subarray (L, R) represents a slice of the original array starting from the element indexed L to the element indexed R.

 Example:
 Input: [1,2,1,2,1,2,1]
 Output: True
 Explanation:
 i = 1, j = 3, k = 5.
 sum(0, i - 1) = sum(0, 0) = 1
 sum(i + 1, j - 1) = sum(2, 2) = 1
 sum(j + 1, k - 1) = sum(4, 4) = 1
 sum(k + 1, n - 1) = sum(6, 6) = 1

 Note:
 1 <= n <= 2000.
 Elements in the given array will be in range [-1,000,000, 1,000,000].
 */
public class _548 {

    public boolean splitArray(int[] nums) {
        int len = nums.length;
        if (len < 7) {
            return false;
        }
        int[] sum = new int[len];
        sum[0] = nums[0];
        for (int i = 1; i < len; i++) {
            sum[i] = sum[i - 1] + nums[i];
        }
        for (int j = 3; j < len - 3; j++) {
            Set<Integer> set = new HashSet<>();
            for (int i = 1; i < j - 1; i++) {
                if (sum[i - 1] == sum[j - 1] - sum[i]) {
                    /**this is sum(0, i-1) and sum(i+1, j-1)*/
                    set.add(sum[i - 1]);
                }
            }
            for (int k = j + 2; k < len - 1; k++) {
                if (sum[k - 1] - sum[j] == sum[len - 1] - sum[k] && set.contains(sum[k - 1] - sum[j])) {
                    /**this is sum(j+1, k-1) and sum(k+1, len-1)*/
                    return true;
                }
            }
        }
        return false;
    }

}

refactor 548