# Find the minimum value from an array associated with another array

Given an integer array A[] and a character array B[] of equal lengths where every character of the array is from the set {‘a’, ‘b’, ‘c’}. Elements of both the arrays are associated with each other i.e. value of B[i] is linked to A[i] for all valid values of i. The task is to find the value min(a + b, c).

Examples:

Input: A[] = {3, 6, 4, 5, 6}, B[] = {‘a’, ‘c’, ‘b’, ‘b’, ‘a’}
Output: 6

Input: A[] = {4, 2, 6, 2, 3}, B[] = {‘b’, ‘a’, ‘c’, ‘a’, ‘b’}
Output: 5

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: In order to minimise the required value, the values of a, b and c has to be minimised. So, traverse the array and find the minimum values of a, b and c associated with these characters in the integer array and finally return min(a + b, c).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to get the minimum required value ` `int` `getMinimum(``int` `A[], ``char` `B[], ``int` `n) ` `{ ` ` `  `    ``// To store the minimum values ` `    ``// of 'a', 'b' and 'c' ` `    ``int` `minA = INT_MAX; ` `    ``int` `minB = INT_MAX; ` `    ``int` `minC = INT_MAX; ` ` `  `    ``// For every value of A[] ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``switch` `(B[i]) { ` ` `  `        ``// Update the minimum values of 'a', ` `        ``// 'b' and 'c' ` `        ``case` `'a'``: ` `            ``minA = min(A[i], minA); ` `            ``break``; ` `        ``case` `'b'``: ` `            ``minB = min(A[i], minB); ` `            ``break``; ` `        ``case` `'c'``: ` `            ``minC = min(A[i], minC); ` `            ``break``; ` `        ``} ` `    ``} ` ` `  `    ``// Return the minimum required value ` `    ``return` `min(minA + minB, minC); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `A[] = { 4, 2, 6, 2, 3 }; ` `    ``char` `B[] = { ``'b'``, ``'a'``, ``'c'``, ``'a'``, ``'b'` `}; ` ` `  `    ``int` `n = ``sizeof``(A) / ``sizeof``(A[0]); ` ` `  `    ``cout << getMinimum(A, B, n); ` `} `

## Java

 `// Java implementation of the above approach ` `class` `GFG ` `{ ` ` `  `// Function to get the minimum required value ` `static` `int` `getMinimum(``int` `A[], ``char` `B[], ``int` `n) ` `{ ` ` `  `    ``// To store the minimum values ` `    ``// of 'a', 'b' and 'c' ` `    ``int` `minA = Integer.MAX_VALUE; ` `    ``int` `minB = Integer.MAX_VALUE; ` `    ``int` `minC = Integer.MAX_VALUE; ` ` `  `    ``// For every value of A[] ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `        ``switch` `(B[i])  ` `        ``{ ` ` `  `            ``// Update the minimum values of 'a', ` `            ``// 'b' and 'c' ` `            ``case` `'a'``: ` `                ``minA = Math.min(A[i], minA); ` `                ``break``; ` `                 `  `            ``case` `'b'``: ` `                ``minB = Math.min(A[i], minB); ` `                ``break``; ` `                 `  `            ``case` `'c'``: ` `                ``minC = Math.min(A[i], minC); ` `                ``break``; ` `        ``} ` `    ``} ` ` `  `    ``// Return the minimum required value ` `    ``return` `Math.min(minA + minB, minC); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int` `A[] = { ``4``, ``2``, ``6``, ``2``, ``3` `}; ` `    ``char` `B[] = { ``'b'``, ``'a'``, ``'c'``, ``'a'``, ``'b'` `}; ` ` `  `    ``int` `n = A.length; ` ` `  `    ``System.out.println(getMinimum(A, B, n)); ` `} ` `}  ` ` `  `// This code is contributed by Rajput-Ji `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to get the minimum required value ` `def` `getMinimum(A, B, n): ` ` `  `    ``# To store the minimum values ` `    ``# of 'a', 'b' and 'c' ` `    ``minA ``=` `float``(``'inf'``); ` `    ``minB ``=` `float``(``'inf'``); ` `    ``minC ``=` `float``(``'inf'``); ` ` `  `    ``# For every value of A[] ` `    ``for` `i ``in` `range``(n): ` `        ``if` `B[i]``=``=``'a'``: ` `            ``minA ``=` `min``(A[i], minA) ` `        ``if` `B[i]``=``=``'b'``: ` `            ``minB ``=` `min``(A[i], minB) ` `        ``if` `B[i]``=``=``'c'``: ` `            ``minB ``=` `min``(A[i], minC) ` ` `  `    ``# Return the minimum required value ` `    ``return` `min``(minA ``+` `minB, minC) ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``A ``=` `[ ``4``, ``2``, ``6``, ``2``, ``3` `] ` `    ``B ``=` `[ ``'b'``, ``'a'``, ``'c'``, ``'a'``, ``'b'` `] ` `    ``n ``=` `len``(A); ` ` `  `    ``print``(getMinimum(A, B, n)) ` ` `  `# This code is contributed by Ashutosh450 `

## C#

 `// C# implementation of the above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `    ``// Function to get the minimum required value ` `    ``static` `int` `getMinimum(``int` `[]A, ``char` `[]B, ``int` `n) ` `    ``{ ` `     `  `        ``// To store the minimum values ` `        ``// of 'a', 'b' and 'c' ` `        ``int` `minA = ``int``.MaxValue; ` `        ``int` `minB = ``int``.MaxValue; ` `        ``int` `minC = ``int``.MaxValue; ` `     `  `        ``// For every value of A[] ` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `            ``switch` `(B[i])  ` `            ``{ ` `     `  `                ``// Update the minimum values of 'a', ` `                ``// 'b' and 'c' ` `                ``case` `'a'``: ` `                    ``minA = Math.Min(A[i], minA); ` `                    ``break``; ` `                     `  `                ``case` `'b'``: ` `                    ``minB = Math.Min(A[i], minB); ` `                    ``break``; ` `                     `  `                ``case` `'c'``: ` `                    ``minC = Math.Min(A[i], minC); ` `                    ``break``; ` `            ``} ` `        ``} ` `     `  `        ``// Return the minimum required value ` `        ``return` `Math.Min(minA + minB, minC); ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``int` `[]A = { 4, 2, 6, 2, 3 }; ` `        ``char` `[]B = { ``'b'``, ``'a'``, ``'c'``, ``'a'``, ``'b'` `}; ` `     `  `        ``int` `n = A.Length; ` `     `  `        ``Console.WriteLine(getMinimum(A, B, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```5
```

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