A>=B Arrays problems

Problem Statement

Amy has an array A, of N integers. Initially all the elements of array are 0, i.e., Ai = 0 (1<=i<=N).

She also has an array B of N integers.

In one operation she can chose an index K, (1<=K<=N) and increase each element of array A, by 1 except AK. (1-based indexing).

For example, N = 4, A = [0, 0, 0, 0], she can choose K = 2 and increase all the elements of array except A2. So, the array becomes A = [1, 0, 1,1]

Find minimum number of operations required for her to make each element of array A, greater than corresponding element in array B, i.e., for each i (1<=i<=N) make Ai>=Bi.

You are given T independent test cases.

Constraints

• 1 <= T <= 5
• 3 <= N <= 105
• 1 <= Bi < = 109

All input values are integers.

Input Format

• First-line contains T.
• First line of each test case consists of a single integer N.
• Second line of each test case consists of N space separated integers denoting the array B.

Output Format

Print in a newline for each test case a single integer denoting the minimum number of operations required for her to make each element of array A, greater than corresponding element in array B, i.e., for each i (1<=i<=N) make Ai>=Bi.

Sample Input 1

1
3
2 1 1

Sample Output 1

2

Explanation of Sample 1

• Initially A = [0, 0, 0]
• In 1st operation, she choses K = 2, so A = [1, 0, 1]
• In 2nd operation she choses K = 3, so A = [ 2, 1, 1]
• Now for each i (1<=i<=3), Ai = Bi

I tried but not get 100% testcases success , help me to solve it

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine().trim());

while (t-- > 0) {
    int n = Integer.parseInt(br.readLine().trim());

    String[] s = br.readLine().trim().split(" ");
    int[] a = new int[n];
    for (int i = 0; i < n; i++) {
        a[i] = Integer.parseInt(s[i]);

    }
    int mi = 0;
    int v = 0;
    int ans = 0;
    int max = a[0];

    for (int i = 1; i < n; i++) {
        if (a[i] > max) max = a[i];
    }
    System.out.println(max);
    Arrays.sort(a);
    for (int i = 0; i < n - 2; i = i + 2) {
        if ((a[i] == max)) a[i] = 0;
        if (a[i + 1] == max) a[i + 1] = 0;
        mi = mi + Math.abs(a[i] - a[i + 1]);
    }
    System.out.println(mi);
    for (int i = 0; i < n; i++) {
        if (a[i] == 0) a[i] = max;
        v = v + (max - a[i]);
    }
    System.out.println(v);

    if (v >= max) ans = max;
    else if (mi == 0) ans = max + mi + 1;
    else ans = max + mi;

    System.out.println(ans);
}