Equal Sum Arrays With Minimum Number of Operations LeetCode solution in Java

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class Solution {
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  public int minOperations(int[] nums1, int[] nums2) {
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    int m = nums1.length, n = nums2.length;
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    if (m > 6 * n || n > 6 * m) return -1;
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    int sum1 = 0, sum2 = 0;
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    for (int i : nums1) sum1 += i;
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    for (int i : nums2) sum2 += i;
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    int diff = sum1 - sum2;
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    if (diff == 0) return 0;
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    return (diff > 0 ? helper(nums1, nums2, diff) : helper(nums2, nums1, -diff));
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  }
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  private int helper(int[] nums1, int[] nums2, int diff) {
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    // count[i] : frequency of numbers that can reduce the diff by i
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    int[] count = new int[6];
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    for (int num : nums1) count[num - 1]++;
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    for (int num : nums2) count[6 - num]++;
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    int res = 0;
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    for (int i = 5; i > 0; i--) {
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      int c = Math.min(count[i], diff / i + (diff % i == 0 ? 0 : 1));
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      res += c;
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      diff -= c * i;
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      if (diff <= 0) break;
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    }
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    return res;
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  }
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}