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# Minimize sum of changes in Array such that ratio of new element to array sum is at most p:q

Given an array A of n integers and two integers p and q, the task is to increase some (or all) of the values in array A in such a way that ratio of each element (after first element) with respect to the total sum of all elements before the current element remains less than or equal to p/q. Return the minimum sum of changes to be done.

Examples:

Input: n = 4, A = {20100, 1, 202, 202}, p = 1, q = 100
Output: 99
Explanation: 50 is added to the first element and 49 to the second element, so that the resulting array becomes {20150, 50, 202, 202}.
50/20150 <= 1/100
202/(20150+50) <=1/100
202/(20150+50+202)<=1/100
Therefore, the condition is satisfied for all the elements and hence 50+49=99 would be the answer.
There are other answers possible as well, but we need to find out the minimum possible sum.

Input: n = 3, A = {1, 1, 1}, p = 100, q = 100
Output: 0

Approach: The problem can be easily solved using the concepts of prefix sum and binary search

• It can be clearly observed that
• if the condition stated in the problem is achieved for a certain sum of changes (say S),
• then it is always possible to achieve the condition for all the numbers greater than S and
• cannot be achieved for all numbers less than S.
• So, we can apply binary search to find the answer.
• Also, it must be carefully observed that instead of distributing S(sum of changes) over different elements, if we just add it to first element, that would not affect the answer.

For example, in the first example above, if 99 is added to the first element, the resultant array will still meet the required condition.

Follow the steps below to solve the problem –

• Make a prefix sum array of the given array.
• Using binary search on range 0 to INT_MAX find the minimum possible answer.

NOTE: Division may lead to overlapping values and errors.

So, instead of comparison like (a/b)<=(c/d), we will do (a*d)<=(b*c).

Below is the implementation of the above approach:

## C++

 `// C++ program for find minimum``// sum of changes in an array``#include ``using` `namespace` `std;` `// function to check if the candidate sum``// satisfies the condition of the problem``bool` `isValid(``int` `candidate, ``int` `pre[], ``int` `n, ``int` `A[],``             ``int` `p, ``int` `q)``{``    ``// flag variable to check wrong answer``    ``bool` `flag = ``true``;``    ``for` `(``int` `i = 1; i < n; i++) {` `        ``// Now for each element, we are checking``        ``// if its ratio with sum of all previous``        ``// elements + candidate is greater than p/q.``        ``// If so, we will return false.``        ``int` `curr_sum = pre[i - 1] + candidate;``        ``if` `(A[i] * q > p * curr_sum) {``            ``flag = ``false``;``            ``break``;``        ``}``        ``// comparing like A[i]/(curr_sum)>p/q``        ``// will be error prone.``    ``}``    ``return` `flag;``}` `int` `solve(``int` `n, ``int` `A[], ``int` `p, ``int` `q)``{` `    ``// declaring and constructing``    ``// prefix sum array``    ``int` `pre[n];``    ``pre[0] = A[0];``    ``for` `(``int` `i = 1; i < n; i++) {``        ``pre[i] = A[i] + pre[i - 1];``    ``}` `    ``// setting lower and upper bound for``    ``// binary search``    ``int` `lo = 0, hi = INT_MAX, ans = INT_MAX;` `    ``// since minimum answer is needed,``    ``// so it is initialized with INT_MAX``    ``while` `(lo <= hi) {` `        ``// calculating mid by using (lo+hi)/2``        ``// may overflow in certain cases``        ``int` `mid = lo + (hi - lo) / 2;` `        ``// checking if required ratio would be``        ``// achieved by all elements if "mid" is``        ``// considered as answer``        ``if` `(isValid(mid, pre, n, A, p, q)) {``            ``ans = mid;``            ``hi = mid - 1;``        ``}``        ``else` `{``            ``lo = mid + 1;``        ``}``    ``}``    ``return` `ans;``}` `// Driver Function``int` `main()``{``    ``int` `n, p, q;``    ``n = 4, p = 1, q = 100;``    ``int` `A[] = { 20100, 1, 202, 202 };` `    ``// printing the required answer``    ``cout << solve(n, A, p, q) << endl;``}`

## Java

 `// Java program for find minimum``// sum of changes in an arraykage whatever //do not write package name here */``import` `java.io.*;` `class` `GFG {` `  ``// function to check if the candidate sum``  ``// satisfies the condition of the problem``  ``static` `Boolean isValid(``int` `candidate, ``int` `pre[], ``int` `n, ``int` `A[],``                         ``int` `p, ``int` `q)``  ``{``    ` `    ``// flag variable to check wrong answer``    ``Boolean flag = ``true``;``    ``for` `(``int` `i = ``1``; i < n; i++) {` `      ``// Now for each element, we are checking``      ``// if its ratio with sum of all previous``      ``// elements + candidate is greater than p/q.``      ``// If so, we will return false.``      ``int` `curr_sum = pre[i - ``1``] + candidate;``      ``if` `(A[i] * q > p * curr_sum) {``        ``flag = ``false``;``        ``break``;``      ``}``      ``// comparing like A[i]/(curr_sum)>p/q``      ``// will be error prone.``    ``}``    ``return` `flag;``  ``}` `  ``static` `int` `solve(``int` `n, ``int` `A[], ``int` `p, ``int` `q)``  ``{` `    ``// declaring and constructing``    ``// prefix sum array``    ``int` `pre[] = ``new` `int``[n];``    ``pre[``0``] = A[``0``];``    ``for` `(``int` `i = ``1``; i < n; i++) {``      ``pre[i] = A[i] + pre[i - ``1``];``    ``}` `    ``// setting lower and upper bound for``    ``// binary search``    ``int` `lo = ``0``, hi = Integer.MAX_VALUE, ans = Integer.MAX_VALUE;` `    ``// since minimum answer is needed,``    ``// so it is initialized with INT_MAX``    ``while` `(lo <= hi) {` `      ``// calculating mid by using (lo+hi)/2``      ``// may overflow in certain cases``      ``int` `mid = lo + (hi - lo) / ``2``;` `      ``// checking if required ratio would be``      ``// achieved by all elements if "mid" is``      ``// considered as answer``      ``if` `(isValid(mid, pre, n, A, p, q)) {``        ``ans = mid;``        ``hi = mid - ``1``;``      ``}``      ``else` `{``        ``lo = mid + ``1``;``      ``}``    ``}``    ``return` `ans;``  ``}` `  ``// Driver Function``  ``public` `static` `void` `main (String[] args)``  ``{``    ``int` `n = ``4``, p = ``1``, q = ``100``;``    ``int` `A[] = { ``20100``, ``1``, ``202``, ``202` `};` `    ``// printing the required answer``    ``System.out.println(solve(n, A, p, q));``  ``}``}` `// This code is contributed by hrithikgarg03188.`

## Python3

 `# Python code for the above approach``import` `sys` `# function to check if the candidate sum``# satisfies the condition of the problem``def` `isValid(candidate, pre, n, A, p, q) :``    ` `    ``# flag variable to check wrong answer``    ``flag ``=` `True``    ``for` `i ``in` `range``(``1``, n) :`` ` `        ``# Now for each element, we are checking``        ``# if its ratio with sum of all previous``        ``# elements + candidate is greater than p/q.``        ``# If so, we will return false.``        ``curr_sum ``=` `pre[i ``-` `1``] ``+` `candidate``        ``if` `(A[i] ``*` `q > p ``*` `curr_sum) :``            ``flag ``=` `False``            ``break``        ` `        ``# comparing like A[i]/(curr_sum)>p/q``        ``# will be error prone.``    ` `    ``return` `flag` ` ` `def` `solve(n, A, p, q) :`` ` `    ``# declaring and constructing``    ``# prefix sum array``    ``pre ``=` `[``0``] ``*` `100``    ``pre[``0``] ``=` `A[``0``]``    ``for` `i ``in` `range``(``1``, n) :``        ``pre[i] ``=` `A[i] ``+` `pre[i ``-` `1``]``    ` ` ` `    ``# setting lower and upper bound for``    ``# binary search``    ``lo ``=` `0``    ``hi ``=` `sys.maxsize``    ``ans ``=` `sys.maxsize`` ` `    ``# since minimum answer is needed,``    ``# so it is initialized with INT_MAX``    ``while` `(lo <``=` `hi) :`` ` `        ``# calculating mid by using (lo+hi)/2``        ``# may overflow in certain cases``        ``mid ``=` `lo ``+` `(hi ``-` `lo) ``/``/` `2`` ` `        ``# checking if required ratio would be``        ``# achieved by all elements if "mid" is``        ``# considered as answer``        ``if` `(isValid(mid, pre, n, A, p, q)) :``            ``ans ``=` `mid``            ``hi ``=` `mid ``-` `1``        ` `        ``else` `:``            ``lo ``=` `mid ``+` `1``        ` `    ``return` `ans`` ` `# Driver Function``n ``=` `4``p ``=` `1``q ``=` `100``A ``=` `[ ``20100``, ``1``, ``202``, ``202` `]`` ` `# printing the required answer``print``(solve(n, A, p, q))` `# This code is contributed by code_hunt.`

## C#

 `// C# program for find minimum``// sum of changes in an array``using` `System;``class` `GFG``{` `  ``// function to check if the candidate sum``  ``// satisfies the condition of the problem``  ``static` `bool` `isValid(``int` `candidate, ``int` `[]pre, ``int` `n, ``int` `[]A,``                      ``int` `p, ``int` `q)``  ``{` `    ``// flag variable to check wrong answer``    ``bool` `flag = ``true``;``    ``for` `(``int` `i = 1; i < n; i++) {` `      ``// Now for each element, we are checking``      ``// if its ratio with sum of all previous``      ``// elements + candidate is greater than p/q.``      ``// If so, we will return false.``      ``int` `curr_sum = pre[i - 1] + candidate;``      ``if` `(A[i] * q > p * curr_sum) {``        ``flag = ``false``;``        ``break``;``      ``}``      ``// comparing like A[i]/(curr_sum)>p/q``      ``// will be error prone.``    ``}``    ``return` `flag;``  ``}` `  ``static` `int` `solve(``int` `n, ``int` `[]A, ``int` `p, ``int` `q)``  ``{` `    ``// declaring and constructing``    ``// prefix sum array``    ``int` `[]pre = ``new` `int``[n];``    ``pre[0] = A[0];``    ``for` `(``int` `i = 1; i < n; i++) {``      ``pre[i] = A[i] + pre[i - 1];``    ``}` `    ``// setting lower and upper bound for``    ``// binary search``    ``int` `lo = 0, hi = Int32.MaxValue, ans = Int32.MaxValue;` `    ``// since minimum answer is needed,``    ``// so it is initialized with INT_MAX``    ``while` `(lo <= hi) {` `      ``// calculating mid by using (lo+hi)/2``      ``// may overflow in certain cases``      ``int` `mid = lo + (hi - lo) / 2;` `      ``// checking if required ratio would be``      ``// achieved by all elements if "mid" is``      ``// considered as answer``      ``if` `(isValid(mid, pre, n, A, p, q)) {``        ``ans = mid;``        ``hi = mid - 1;``      ``}``      ``else` `{``        ``lo = mid + 1;``      ``}``    ``}``    ``return` `ans;``  ``}` `  ``// Driver Function``  ``public` `static` `void` `Main()``  ``{``    ``int` `n = 4, p = 1, q = 100;``    ``int` `[]A = { 20100, 1, 202, 202 };` `    ``// printing the required answer``    ``Console.WriteLine(solve(n, A, p, q));``  ``}``}` `// This code is contributed by Samim Hossain Mondal.`

## Javascript

 ``

Output

`99`

Time Complexity: O(n log(INT_MAX))
Auxiliary space: O(n)

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