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# Most subtraction of Array values from Ok to make it 0 or minimal

Given two arrays A[] and B[] (B[i] < A[i]) of measurement N every and an integer Ok, the duty is to carry out the operation of (Ok – A[i] + B[i]) the utmost variety of instances such that Ok turns into 0 or as minimal as doable.

Be aware: At no second an operation will likely be carried out such that Ok turns into damaging i.e. the operation can’t be carried out if Ok – A[i] is damaging and any index might be chosen as many instances as doable.

Examples:

Enter: N = 3, Ok = 3, A[] = {2, 3, 4}, B[] = {1, 1, 1}
Output: 2
Clarification: For optimum turns,
Select A and B, so  Ok = 3 – 2 + 1 = 2.
Select A and B, so  Ok = 2 – 2 + 1 = 1.
No different operation is feasible. So max turns = 2.

Enter: N = 2, Ok = 10, A[] = {5, 4} B[] ={1, 2}
Output: 4

Method: The issue might be solved primarily based on the next thought:

For performing the operation most variety of instances, all the time preserve selecting the indices (say i) in order that A[i] – B[i] is minimal.

Say such worth is A[i] and B[i], so the utmost carried out operations for that index are (Ok – B[i]) / (A[i] – B[i]) [Here K – B[i] to keep away from the final operation the place Ok-A[i] is damaging]

Comply with the steps talked about beneath to implement the concept:

• Create an array (say v) to retailer the values within the type of (A[i] – B[i], A[i]).
• Type the array in growing order on the idea of (A[i] – B[i]).
• Loop from i = 0 to N-1:
• If Ok – second worth of v[i] will not be damaging then use the above components to search out its contribution (say x).
• Increment the depend by x.
• Return the ultimate depend as the reply.

Under is the implementation of the above strategy.

## C++

 ` `  `#embrace ` `utilizing` `namespace` `std;` ` `  `int` `MakeitMin(``int` `n, ``int` `okay, ``int` `a[], ``int` `b[])` `{` `    ` `    ``vector > v(n);` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``v[i] = { a[i] - b[i], a[i] };` ` `  `    ` `    ``kind(v.start(), v.finish());` `    ``int` `ans = 0;` `    ``for` `(``int` `i = 0; i < n; i++) {` ` `  `        ` `        ``if` `(v[i].second > okay)` `            ``proceed``;` ` `  `        ` `        ` `        ` `        ``int` `diff` `            ``= (okay - (v[i].second - v[i].first)) / v[i].first;` `        ``ans += diff;` ` `  `        ` `        ``okay -= (diff * v[i].first);` `    ``}` `    ``return` `ans;` `}` ` `  `int` `foremost()` `{` `    ``int` `N = 2, Ok = 10;` `    ``int` `A[] = { 5, 4 };` `    ``int` `B[] = { 1, 2 };` `    ``cout << MakeitMin(N, Ok, A, B);` `    ``return` `0;` `}`

Time Complexity: O(N * logN)
Auxiliary Area: O(N)

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