# Maximize product of digit sum of consecutive pairs in a subsequence of length K

Given an array of integers arr[], the task is to maximize the product of the digit sum of every consecutive pairs in a subsequence of length K.

Note: K is always even because pairs will be formed on the even length.

Examples:

Input: arr[] = {2, 100, 99, 3, 16}, K = 4
Output: 128
The optimal subsequence of length 4 is [2, 100, 99, 16]
Sum of digits = 2, 2, 18 and 7 respectively.
So Product of digit sums in pairs = 2 * 1 + 18 * 7 = 2 + 126 = 128, which is maximum.

Input: arr[] = {10, 5, 9, 101, 24, 2, 20, 14}, K = 6
Output: 69
The optimal subsequence of length 6 = [10, 5, 9, 24, 2, 14]
Sum of digits = 1, 5, 9, 6, 2 and 5 respectively.
So Product of digit sums in pairs = 1 * 5 + 9 * 6 + 2 * 5 = 5 + 54 + 10 = 69, which is maximum.

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

Approach: The idea is to use Dynamic Programming. As we need to find pairs in the array by including or excluding some of the elements from the array to form a subsequence. So let DP[i][j][k] be our dp array which stores the maximum product of the sum of digits of the elements upto index i having length j and last element as K.

Observations:

• Odd Length: While choosing the even length subsequence, when current length of the choosen subsequence is odd, then the pair for the last element is to be chosen. Therefore, we have to compute the product of sum of the last and current element and we recur by keeping last as 0 in next call for even length. • Even Length: While choosing the odd length subsequence, when current length of the choosen subsequence is even, then we have to just select the first element of the pair. Therefore, we just select the current element as last element for the next recursive call and search for second element for this pair in next recursive call. • Exclude Element: Another option for the current element is to exclude the current element and choose the elements further. Below is the implementation of above approach:

## C++

 `// C++ implementation to find the ` `// maximum product of the digit ` `// sum of the consecutive pairs of ` `// the subsequence of the length K ` ` `  `#include ` `using` `namespace` `std; ` ` `  `const` `int` `MAX = 100; ` `int` `dp[MAX][MAX]; ` ` `  `// Function to find the product ` `// of two numbers digit sum ` `// in the pair ` `int` `productDigitSum(``int` `x, ``int` `y) ` `{ ` `    ``int` `sumx = 0; ` ` `  `    ``// Loop to find the digits of ` `    ``// the number ` `    ``while` `(x) { ` `        ``sumx += (x % 10); ` `        ``x /= 10; ` `    ``} ` `    ``int` `sumy = 0; ` ` `  `    ``// Loop to find the digits ` `    ``// of other number ` `    ``while` `(y) { ` `        ``sumy += (y % 10); ` `        ``y /= 10; ` `    ``} ` `    ``return` `(sumx * sumy); ` `} ` ` `  `// Function to find the subsequence ` `// of the length K ` `int` `solve(``int` `arr[], ``int` `i, ``int` `len, ` `          ``int` `prev, ``int` `n, ``int` `k) ` `{ ` `    ``// Base Case ` `    ``if` `(len == k) ` `        ``return` `0; ` ` `  `    ``// Condition when we didn't reach ` `    ``// the length K, but ran out of ` `    ``// elements of the array ` `    ``if` `(i == n) ` `        ``return` `INT_MIN; ` ` `  `    ``// Condition if already calculated ` `    ``if` `(dp[i][len][prev]) ` `        ``return` `dp[i][len][prev]; ` ` `  `    ``int` `inc = 0, exc = 0; ` ` `  `    ``// If length upto this point is odd ` `    ``if` `(len & 1) { ` ` `  `        ``// If length is odd, it means we need ` `        ``// second element of this current pair, ` `        ``// calculate the product of digit sum of ` `        ``// current and previous element and recur ` `        ``// by moving towards next index ` `        ``inc = productDigitSum(arr[prev], ` `                              ``arr[i]) ` `              ``+ solve(arr, i + 1, ` `                      ``len + 1, 0, n, k); ` `    ``} ` ` `  `    ``// If length upto this point is even ` `    ``else` `{ ` `        ``inc = solve(arr, i + 1, len + 1, i, n, k); ` `    ``} ` ` `  `    ``// Exclude this current element ` `    ``// and recur for next elements. ` `    ``exc = solve(arr, i + 1, len, prev, n, k); ` ` `  `    ``// return by memoizing it, by selecting ` `    ``// the maximum among two choices. ` `    ``return` `dp[i][len][prev] = max(inc, exc); ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 10, 5, 9, 101, 24, 2, 20, 14 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `k = 6; ` `    ``cout << solve(arr, 0, 0, 0, n, k); ` `} `

## Java

 `// Java implementation to find the ` `// maximum product of the digit ` `// sum of the consecutive pairs of ` `// the subsequence of the length K ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `     `  `static` `int` `MAX = ``100``; ` `static` `int` `dp[][][] = ``new` `int``[``1000``][MAX][MAX]; ` ` `  `// Function to find the product ` `// of two numbers digit sum ` `// in the pair ` `static` `int` `productDigitSum(``int` `x, ``int` `y) ` `{ ` `    ``int` `sumx = ``0``; ` ` `  `    ``// Loop to find the digits  ` `    ``// of the number ` `    ``while` `(x > ``0``)  ` `    ``{ ` `        ``sumx += (x % ``10``); ` `        ``x /= ``10``; ` `    ``} ` `    ``int` `sumy = ``0``; ` ` `  `    ``// Loop to find the digits ` `    ``// of other number ` `    ``while` `(y > ``0``) ` `    ``{ ` `        ``sumy += (y % ``10``); ` `        ``y /= ``10``; ` `    ``} ` `    ``return` `(sumx * sumy); ` `} ` ` `  `// Function to find the subsequence ` `// of the length K ` `static` `int` `solve(``int` `arr[], ``int` `i, ``int` `len, ` `                  ``int` `prev, ``int` `n, ``int` `k) ` `{ ` `     `  `    ``// Base Case ` `    ``if` `(len == k) ` `        ``return` `0``; ` ` `  `    ``// Condition when we didn't reach ` `    ``// the length K, but ran out of ` `    ``// elements of the array ` `    ``if` `(i == n) ` `        ``return` `Integer.MIN_VALUE; ` ` `  `    ``// Condition if already calculated ` `    ``if` `(dp[i][len][prev] != ``0``) ` `        ``return` `dp[i][len][prev]; ` ` `  `    ``int` `inc = ``0``, exc = ``0``; ` ` `  `    ``// If length upto this point is odd ` `    ``if` `((len & ``1``) != ``0``) ` `    ``{ ` ` `  `        ``// If length is odd, it means we need ` `        ``// second element of this current pair, ` `        ``// calculate the product of digit sum of ` `        ``// current and previous element and recur ` `        ``// by moving towards next index ` `        ``inc = (productDigitSum(arr[prev], arr[i]) +  ` `               ``solve(arr, i + ``1``, len + ``1``, ``0``, n, k)); ` `    ``} ` ` `  `    ``// If length upto this point is even ` `    ``else`  `    ``{ ` `        ``inc = solve(arr, i + ``1``, len + ``1``, i, n, k); ` `    ``} ` ` `  `    ``// Exclude this current element ` `    ``// and recur for next elements. ` `    ``exc = solve(arr, i + ``1``, len, prev, n, k); ` ` `  `    ``// Return by memoizing it, by selecting ` `    ``// the maximum among two choices. ` `    ``return` `dp[i][len][prev] = Math.max(inc, exc); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String []args) ` `{ ` `    ``int` `arr[] = { ``10``, ``5``, ``9``, ``101``, ``24``, ``2``, ``20``, ``14` `}; ` `    ``int` `n = arr.length; ` `    ``int` `k = ``6``; ` `     `  `    ``System.out.print(solve(arr, ``0``, ``0``, ``0``, n, k)); ` `} ` `} ` ` `  `// This code is contributed by chitranayal `

## Python3

 `# Python3 implementation to find the  ` `# maximum product of the digit  ` `# sum of the consecutive pairs of  ` `# the subsequence of the length K  ` `import` `sys ` ` `  `MAX` `=` `100` `dp ``=` `[] ` ` `  `for` `i ``in` `range``(``1000``): ` `    ``temp1 ``=` `[] ` `    ``for` `j ``in` `range``(``MAX``): ` `        ``temp2 ``=` `[] ` `         `  `        ``for` `k ``in` `range``(``MAX``): ` `            ``temp2.append(``0``) ` `        ``temp1.append(temp2) ` `         `  `    ``dp.append(temp1) ` ` `  `# Function to find the product  ` `# of two numbers digit sum  ` `# in the pair  ` `def` `productDigitSum(x, y): ` `    ``sumx ``=` `0` `     `  `    ``# Loop to find the digits of  ` `    ``# the number ` `    ``while` `x: ` `        ``sumx ``+``=` `x ``%` `10` `        ``x ``=` `x ``/``/` `10` `         `  `    ``sumy ``=` `0` `     `  `    ``# Loop to find the digits  ` `    ``# of other number  ` `    ``while` `y: ` `        ``sumy ``+``=` `y ``%` `10` `        ``y ``=` `y ``/``/` `10` `     `  `    ``return` `sumx ``*` `sumy ` ` `  `# Function to find the subsequence  ` `# of the length K  ` `def` `solve(arr, i, ``len``, prev, n, k): ` `     `  `    ``# Base case ` `    ``if` `len` `=``=` `k: ` `        ``return` `0` `     `  `    ``# Condition when we didn't reach  ` `    ``# the length K, but ran out of  ` `    ``# elements of the array  ` `    ``if` `i ``=``=` `n: ` `        ``return` `-``sys.maxsize ``-` `1` `     `  `    ``# Condition if already calculated ` `    ``if` `dp[i][``len``][prev]: ` `        ``return` `dp[i][``len``][prev] ` `     `  `    ``# If length upto this point is odd ` `    ``if` `len` `& ``1``: ` `         `  `        ``# If length is odd, it means we need  ` `        ``# second element of this current pair,  ` `        ``# calculate the product of digit sum of  ` `        ``# current and previous element and recur  ` `        ``# by moving towards next index  ` `        ``inc ``=` `(productDigitSum(arr[prev], arr[i]) ``+`  `               ``solve(arr, i ``+` `1``, ``len` `+` `1``, i, n, k)) ` `    ``else``: ` `         `  `        ``# If length upto this point is even ` `        ``inc ``=` `solve(arr, i ``+` `1``, ``len` `+` `1``, i, n, k) ` `     `  `    ``# Exclude this current element  ` `    ``# and recur for next elements.  ` `    ``exc ``=` `solve(arr, i ``+` `1``, ``len``, prev, n, k) ` `     `  `    ``# Return by memoizing it, by selecting  ` `    ``# the maximum among two choices. ` `    ``dp[i][``len``][prev] ``=` `max``(inc, exc) ` `    ``return` `dp[i][``len``][prev] ` ` `  `# Driver code ` `arr ``=` `[ ``10``, ``5``, ``9``, ``101``, ``24``, ``2``, ``20``, ``14` `] ` `n ``=` `len``(arr) ` `k ``=` `6` `print``(solve(arr, ``0``, ``0``, ``0``, n, k)) ` ` `  `# This code is contributed by Shivam Singh `

## C#

 `// C# implementation to find the ` `// maximum product of the digit ` `// sum of the consecutive pairs of ` `// the subsequence of the length K ` `using` `System; ` `class` `GFG{ ` `     `  `static` `int` `MAX = 100; ` `static` `int` `[, ,]dp = ``new` `int``[1000, MAX, MAX]; ` ` `  `// Function to find the product ` `// of two numbers digit sum ` `// in the pair ` `static` `int` `productDigitSum(``int` `x, ``int` `y) ` `{ ` `    ``int` `sumx = 0; ` ` `  `    ``// Loop to find the digits  ` `    ``// of the number ` `    ``while` `(x > 0)  ` `    ``{ ` `        ``sumx += (x % 10); ` `        ``x /= 10; ` `    ``} ` `    ``int` `sumy = 0; ` ` `  `    ``// Loop to find the digits ` `    ``// of other number ` `    ``while` `(y > 0) ` `    ``{ ` `        ``sumy += (y % 10); ` `        ``y /= 10; ` `    ``} ` `    ``return` `(sumx * sumy); ` `} ` ` `  `// Function to find the subsequence ` `// of the length K ` `static` `int` `solve(``int` `[]arr, ``int` `i, ``int` `len, ` `                 ``int` `prev, ``int` `n, ``int` `k) ` `{ ` `     `  `    ``// Base Case ` `    ``if` `(len == k) ` `        ``return` `0; ` ` `  `    ``// Condition when we didn't reach ` `    ``// the length K, but ran out of ` `    ``// elements of the array ` `    ``if` `(i == n) ` `        ``return` `Int32.MinValue; ` ` `  `    ``// Condition if already calculated ` `    ``if` `(dp[i, len, prev] != 0) ` `        ``return` `dp[i, len, prev]; ` ` `  `    ``int` `inc = 0, exc = 0; ` ` `  `    ``// If length upto this point is odd ` `    ``if` `((len & 1) != 0) ` `    ``{ ` ` `  `        ``// If length is odd, it means we need ` `        ``// second element of this current pair, ` `        ``// calculate the product of digit sum of ` `        ``// current and previous element and recur ` `        ``// by moving towards next index ` `        ``inc = (productDigitSum(arr[prev], arr[i]) +  ` `               ``solve(arr, i + 1, len + 1, 0, n, k)); ` `    ``} ` ` `  `    ``// If length upto this point is even ` `    ``else` `    ``{ ` `        ``inc = solve(arr, i + 1, len + 1, i, n, k); ` `    ``} ` ` `  `    ``// Exclude this current element ` `    ``// and recur for next elements. ` `    ``exc = solve(arr, i + 1, len, prev, n, k); ` ` `  `    ``// Return by memoizing it, by selecting ` `    ``// the maximum among two choices. ` `    ``return` `dp[i, len, prev] = Math.Max(inc, exc); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `[]arr = { 10, 5, 9, 101, 24, 2, 20, 14 }; ` `    ``int` `n = arr.Length; ` `    ``int` `k = 6; ` `     `  `    ``Console.Write(solve(arr, 0, 0, 0, n, k)); ` `} ` `} ` ` `  `// This code is contributed by Nidhi_biet `

Output:

```69
```

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