# Minimum Sum of a pair at least K distance apart from an Array

Given an array of integers A[] of size N, the task is to find the minimum sum that can be obtained by any pair of array elements which are at least K indices apart from each other.

Examples:

Input: A[] = {1, 2, 3, 4, 5, 6}, K = 2
Output:
Explanation:
The minimum sum that can be obtained is by adding 1 and 3 that are at a distance of 2.

Input: A[] = {4, 2, 5, 4, 3, 2, 5}, K = 3
Output:
Explanation:
The minimum sum that can be obtained is by adding 2 and 2 that are at a distance of 4.

Naive Approach:
The simplest approach is to solve the problem is to iterate over the indices [i + K, N – 1] for every ith index and find the minimum element, say min. Check if min + A[i] is less than the minimum sum obtained so far and update minimum_sum accordingly. Finally, print the minimum_sum.

Below is the implementation of the above approach:

 `// C++ program to implement ` `// the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the minimum ` `// sum of two elements that ` `// are atleast K distance apart ` `void` `findMinSum(``int` `A[], ``int` `K, ``int` `n) ` `{ ` `    ``int` `minimum_sum = INT_MAX; ` ` `  `    ``// Iterate over the array ` `    ``for``(``int` `i = 0; i < n; i++) ` `    ``{ ` `         `  `        ``// Initialize the min value ` `        ``int` `mini = INT_MAX; ` ` `  `        ``// Iterate from i + k to N ` `        ``for``(``int` `j = i + K; j < n; j++) ` ` `  `            ``// Find the minimum ` `            ``mini = min(mini, A[j]); ` ` `  `        ``if` `(mini == INT_MAX) ` `            ``continue``; ` ` `  `        ``// Update the minimum sum ` `        ``minimum_sum = min(minimum_sum, ` `                          ``A[i] + mini); ` `    ``} ` ` `  `    ``// Print the answer ` `    ``cout << (minimum_sum); ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `A[] = { 4, 2, 5, 4, 3, 2, 5 }; ` `    ``int` `K = 3; ` `    ``int` `n = ``sizeof``(A) / ``sizeof``(A); ` ` `  `    ``findMinSum(A, K, n); ` `    ``return` `0; ` `} ` ` `  `// This code is contributed by chitranayal`

 `// Java Program to implement ` `// the above approach ` ` `  `import` `java.util.*; ` `class` `GFG { ` ` `  `    ``// Function to find the minimum ` `    ``// sum of two elements that ` `    ``// are atleast K distance apart ` `    ``public` `static` `void` `    ``findMinSum(``int` `A[], ``int` `K) ` `    ``{ ` `        ``// Length of the array ` `        ``int` `n = A.length; ` ` `  `        ``int` `minimum_sum ` `            ``= Integer.MAX_VALUE; ` ` `  `        ``// Iterate over the array ` `        ``for` `(``int` `i = ``0``; i < n; i++) { ` ` `  `            ``// Initialize the min value ` `            ``int` `min = Integer.MAX_VALUE; ` ` `  `            ``// Iterate from i + k to N ` `            ``for` `(``int` `j = i + K; j < n; j++) ` ` `  `                ``// Find the minimum ` `                ``min = Math.min(min, A[j]); ` ` `  `            ``if` `(min == Integer.MAX_VALUE) ` `                ``continue``; ` ` `  `            ``// Update the minimum sum ` `            ``minimum_sum = Math.min(minimum_sum, ` `                                   ``A[i] + min); ` `        ``} ` ` `  `        ``// Print the answer ` `        ``System.out.println(minimum_sum); ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `        ``main(String[] args) ` `    ``{ ` ` `  `        ``int` `A[] = { ``4``, ``2``, ``5``, ``4``, ``3``, ``2``, ``5` `}; ` `        ``int` `K = ``3``; ` ` `  `        ``findMinSum(A, K); ` `    ``} ` `} `

 `# Python3 Program to implement ` `# the above approach ` `import` `sys ` ` `  `# Function to find the minimum ` `# sum of two elements that ` `# are atleast K distance apart ` `def` `findMinSum(A, K): ` `   `  `    ``# Length of the array ` `    ``n ``=` `len``(A); ` ` `  `    ``minimum_sum ``=` `sys.maxsize; ` ` `  `    ``# Iterate over the array ` `    ``for` `i ``in` `range``(n): ` ` `  `        ``# Initialize the min value ` `        ``minmum ``=` `sys.maxsize; ` ` `  `        ``# Iterate from i + k to N ` `        ``for` `j ``in` `range``(i ``+` `K, n, ``1``): ` ` `  `            ``# Find the minimum ` `            ``minmum ``=` `min``(minmum, A[j]); ` ` `  `        ``if` `(minmum ``=``=` `sys.maxsize): ` `            ``continue``; ` ` `  `        ``# Update the minimum sum ` `        ``minimum_sum ``=` `min``(minimum_sum, A[i] ``+` `minmum); ` ` `  `    ``# Prthe answer ` `    ``print``(minimum_sum); ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``A ``=` `[``4``, ``2``, ``5``, ``4``, ``3``, ``2``, ``5``]; ` `    ``K ``=` `3``; ` ` `  `    ``findMinSum(A, K); ` ` `  `# This code is contributed by sapnasingh4991`

 `// C# Program to implement ` `// the above approach ` `using` `System; ` `class` `GFG{ ` ` `  `  ``// Function to find the minimum ` `  ``// sum of two elements that ` `  ``// are atleast K distance apart ` `  ``public` `static` `void` `findMinSum(``int` `[]A, ` `                                ``int` `K) ` `  ``{ ` `    ``// Length of the array ` `    ``int` `n = A.Length; ` ` `  `    ``int` `minimum_sum = ``int``.MaxValue; ` ` `  `    ``// Iterate over the array ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` ` `  `      ``// Initialize the min value ` `      ``int` `min = ``int``.MaxValue; ` ` `  `      ``// Iterate from i + k to N ` `      ``for` `(``int` `j = i + K; j < n; j++) ` ` `  `        ``// Find the minimum ` `        ``min = Math.Min(min, A[j]); ` ` `  `      ``if` `(min == ``int``.MaxValue) ` `        ``continue``; ` ` `  `      ``// Update the minimum sum ` `      ``minimum_sum = Math.Min(minimum_sum, ` `                             ``A[i] + min); ` `    ``} ` ` `  `    ``// Print the answer ` `    ``Console.WriteLine(minimum_sum); ` `  ``} ` ` `  `  ``// Driver Code ` `  ``public` `static` `void` `Main(String[] args) ` `  ``{ ` `    ``int` `[]A = { 4, 2, 5, 4, 3, 2, 5 }; ` `    ``int` `K = 3; ` ` `  `    ``findMinSum(A, K); ` `  ``} ` `} ` ` `  `// This code is contributed by Rohit_ranjan`

Output:
```4
```

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach:
The above approach can be optimized using a Suffix Array. Follow the steps below:

• Initialize a suffix array(say suffix[]), where suffix[i] stores the minimum of all the elements from index N-1 to i.
• For any ith index, the minimum element which is K distance apart is stored at index i + K in the suffix array.
• For i ranging from 0 to N – 1, check if A[i] + suffix[i + k] < minimum_sum or not and update minimum_sum accordingly.
• Finally, print minimum_sum as the required answer.

Below is the implementation of the above approach:

 `// C++ Program to implement ` `//the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the minimum ` `// sum of two elements that ` `// are atleast K distance apart ` `void` `findMinSum(``int` `A[], ``int` `K, ``int` `len) ` `{ ` ` `  `  ``// Length of the array ` `  ``int` `n = len; ` `  ``int` `suffix_min[n] = {0}; ` ` `  `  ``suffix_min[n - 1] = A[n - 1]; ` ` `  `  ``// Find the suffix array ` `  ``for` `(``int` `i = n - 2; i >= 0; i--) ` `    ``suffix_min[i] = min(suffix_min[i + 1], A[i]); ` ` `  `  ``int` `min_sum = INT_MAX; ` ` `  `  ``// Iterate in the array ` `  ``for` `(``int` `i = 0; i < n; i++)  ` `  ``{ ` `    ``if` `(i + K < n) ` ` `  `      ``// Update minimum sum ` `      ``min_sum = min(min_sum, A[i] +  ` `                    ``suffix_min[i + K]); ` `  ``} ` ` `  `  ``// Print the answer ` `  ``cout << min_sum; ` `} ` ` `  ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `A[] = { 1, 2, 3, 4, 5, 6 }; ` `    ``int` `K = 2; ` `    ``int` `n = ``sizeof``(A) / ``sizeof``(A); ` `    ``findMinSum(A, K, n); ` `    ``return` `0; ` `} ` ` `  `// This code is contributed by Rohit_ranjan`

 `// Java Program to implement ` `// the above approach ` ` `  `import` `java.util.*; ` `class` `GFG { ` ` `  `    ``// Function to find the minimum ` `    ``// sum of two elements that ` `    ``// are atleast K distance apart ` `    ``public` `static` `void` `    ``findMinSum(``int` `A[], ``int` `K) ` `    ``{ ` ` `  `        ``// Length of the array ` `        ``int` `n = A.length; ` `        ``int` `suffix_min[] = ``new` `int``[n]; ` ` `  `        ``suffix_min[n - ``1``] = A[n - ``1``]; ` ` `  `        ``// Find the suffix array ` `        ``for` `(``int` `i = n - ``2``; i >= ``0``; i--) ` `            ``suffix_min[i] ` `                ``= Math.min(suffix_min[i + ``1``], ` `                           ``A[i]); ` ` `  `        ``int` `min_sum = Integer.MAX_VALUE; ` ` `  `        ``// Iterate in the array ` `        ``for` `(``int` `i = ``0``; i < n; i++) { ` ` `  `            ``if` `(i + K < n) ` ` `  `                ``// Update minimum sum ` `                ``min_sum = Math.min( ` `                    ``min_sum, A[i] ` `                                 ``+ suffix_min[i + K]); ` `        ``} ` ` `  `        ``// Print the answer ` `        ``System.out.println(min_sum); ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `A[] = { ``1``, ``2``, ``3``, ``4``, ``5``, ``6` `}; ` `        ``int` `K = ``2``; ` ` `  `        ``findMinSum(A, K); ` `    ``} ` `} `

 `# Python3 program to implement ` `# the above approach ` `import` `sys ` ` `  `# Function to find the minimum ` `# sum of two elements that ` `# are atleast K distance apart ` `def` `findMinSum(A, K): ` `     `  `    ``# Length of the array ` `    ``n ``=` `len``(A); ` `     `  `    ``suffix_min ``=` `[``0``] ``*` `n; ` `    ``suffix_min[n ``-` `1``] ``=` `A[n ``-` `1``]; ` ` `  `    ``# Find the suffix array ` `    ``for` `i ``in` `range``(n ``-` `2``, ``-``1``, ``-``1``): ` `        ``suffix_min[i] ``=` `min``(suffix_min[i ``+` `1``], A[i]); ` ` `  `    ``min_sum ``=` `sys.maxsize; ` ` `  `    ``# Iterate in the array ` `    ``for` `i ``in` `range``(n): ` `        ``if` `(i ``+` `K < n): ` ` `  `            ``# Update minimum sum ` `            ``min_sum ``=` `min``(min_sum, A[i] ``+`  `                          ``suffix_min[i ``+` `K]); ` ` `  `    ``# Print the answer ` `    ``print``(min_sum); ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `     `  `    ``A ``=` `[ ``1``, ``2``, ``3``, ``4``, ``5``, ``6` `]; ` `    ``K ``=` `2``; ` ` `  `    ``findMinSum(A, K); ` ` `  `# This code is contributed by Amit Katiyar `

 `// C# program to implement ` `// the above approach ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Function to find the minimum ` `// sum of two elements that ` `// are atleast K distance apart ` `public` `static` `void` `findMinSum(``int` `[]A, ``int` `K) ` `{ ` `     `  `    ``// Length of the array ` `    ``int` `n = A.Length; ` `    ``int` `[]suffix_min = ``new` `int``[n]; ` ` `  `    ``suffix_min[n - 1] = A[n - 1]; ` ` `  `    ``// Find the suffix array ` `    ``for``(``int` `i = n - 2; i >= 0; i--) ` `        ``suffix_min[i] = Math.Min(suffix_min[i + 1], ` `                                          ``A[i]); ` ` `  `    ``int` `min_sum = ``int``.MaxValue; ` ` `  `    ``// Iterate in the array ` `    ``for``(``int` `i = 0; i < n; i++)  ` `    ``{ ` `        ``if` `(i + K < n) ` ` `  `            ``// Update minimum sum ` `            ``min_sum = Math.Min(min_sum, A[i] +  ` `                               ``suffix_min[i + K]); ` `    ``} ` ` `  `    ``// Print the answer ` `    ``Console.WriteLine(min_sum); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]A = { 1, 2, 3, 4, 5, 6 }; ` `    ``int` `K = 2; ` ` `  `    ``findMinSum(A, K); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar  `

Output:
```4
```

Time Complexity: O(N)
Auxiliary Space: O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :