# Minimum 1s to lend power to make whole array powerful

Given a binary array and an integer k where every 1 has power and it can lend power to neighbors and itself. A cell with value 1 can lend power within distance <k. Task is to find the minimum number of 1s that need to lend power so that all cells become powerful.

If it is not possible to make whole array powerful, return -1.
Examples:

```Input  : arr[] = {0, 1, 1, 1, 1, 0}
K = 2
Output : 2
arr and arr need to lend

Input  :  arr[] = {1, 0, 1, 0, 0, 1, 0, 0, 0, 1}
K = 3
Output : 3
a, a and a lend to their right side zeros.

Input  :  arr[] = {1, 1, 1}
K = 2
Output : 1
arr need to lend power to all cells.
```

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

A simple solution is to try all possible combinations of 1s, starting with one 1, then two 1s and so on until we find a combination that makes the whole array powerful.

An efficient solution is based on the observation that there should be a 1 within distance k for every cell. We preprocess the array and create an auxiliary array that stores indexes of closest 1 on left side. We traverse through the whole array and find index of the closest left 1 within k distance of current cell. If there is no left 1, we return -1. Else, we increment ans and move to the next cell after k distance.

## C++

 `// C++ program to minimum of number of lendings ` `// by 1s. ` `#include ` `using` `namespace` `std; ` ` `  `int` `minLendings(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``// Create an auxiliary array and fill  ` `    ``// indexes of closest 1s on left side.  ` `    ``int` `leftOne[n]; ` `    ``int` `oneInd = -1; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(arr[i] == 1) ` `            ``oneInd = i; ` `        ``leftOne[i] = oneInd; ` `    ``} ` ` `  `    ``// Traverse the array from left side. If ` `    ``// current element has a 1 on ` `    ``int` `ans = 0; ` `    ``for` `(``int` `i = 0; i < n;) { ` `  `  `        ``// Find index of closest 1 after shift of ` `        ``// k-1 or end whichever is closer. ` `        ``int` `pos = leftOne[min(i + k - 1, n - 1)]; ` ` `  `        ``// If there is no closest 1 within allowed ` `        ``// range.  ` `        ``if` `(pos == -1 || pos + k <= i) ` `            ``return` `-1; ` ` `  `        ``// If a closest 1 is found, check after k ` `        ``// jumps and increment answer. ` `        ``i = pos + k; ` `        ``ans++; ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 0, 1, 1, 1, 1, 0 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `k = 2; ` `    ``cout << minLendings(arr, n, k); ` `    ``return` `0; ` `} `

## Java

 `// Java program to minimum of number of  ` `// lendings by 1s. ` `import` `java.math.*; ` ` `  `class` `GFG { ` ` `  `static` `int` `minLendings(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``// Create an auxiliary array and fill  ` `    ``// indexes of closest 1s on left side.  ` `    ``int` `leftOne[] = ``new` `int``[n]; ` `    ``int` `oneInd = -``1``; ` `    ``for` `(``int` `i = ``0``; i < n; i++) { ` `        ``if` `(arr[i] == ``1``) ` `            ``oneInd = i; ` `        ``leftOne[i] = oneInd; ` `    ``} ` ` `  `    ``// Traverse the array from left side. If ` `    ``// current element has a 1 on ` `    ``int` `ans = ``0``; ` `    ``for` `(``int` `i = ``0``; i < n;) { ` ` `  `        ``// Find index of closest 1 after shift of ` `        ``// k-1 or end whichever is closer. ` `        ``int` `pos = leftOne[Math.min(i + k - ``1``, n - ``1``)]; ` ` `  `        ``// If there is no closest 1 within  ` `        ``// allowed range.  ` `        ``if` `(pos == -``1` `|| pos + k <= i) ` `            ``return` `-``1``; ` ` `  `        ``// If a closest 1 is found, check after k ` `        ``// jumps and increment answer. ` `        ``i = pos + k; ` `        ``ans++; ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``0``, ``1``, ``1``, ``1``, ``1``, ``0` `}; ` `    ``int` `n = arr.length; ` `    ``int` `k = ``2``; ` `    ``System.out.println(minLendings(arr, n, k)); ` `} ` `} ` ` `  `// This code is contributed by Prerna Saini `

## Python3

 `# Python3 program to minimum ` `# of number of lendings by 1s. ` ` `  `def` `minLendings(arr, n, k): ` ` `  `    ``# Create an auxiliary array and fill  ` `    ``# indexes of closest 1s on left side.  ` `    ``leftOne ``=` `[``0` `for` `i ``in` `range``(n ``+` `1``)] ` `    ``oneInd ``=` `-``1` `     `  `    ``for` `i ``in` `range``(n): ` `        ``if` `(arr[i] ``=``=` `1``): ` `            ``oneInd ``=` `i ` `        ``leftOne[i] ``=` `oneInd ` ` `  `    ``# Traverse the array from left side. ` `    ``# If current element has a 1 on ` `    ``ans ``=` `0``; i ``=` `0` `     `  `    ``while``(i < n): ` ` `  `        ``# Find index of closest 1 after shift of ` `        ``# k-1 or end whichever is closer. ` `        ``pos ``=` `leftOne[``min``(i ``+` `k ``-` `1``, n ``-` `1``)] ` ` `  `        ``# If there is no closest 1   ` `        ``# within allowed range.  ` `        ``if` `(pos ``=``=` `-``1` `or` `pos ``+` `k <``=` `i): ` `            ``return` `-``1` ` `  `        ``# If a closest 1 is found, check after k ` `        ``# jumps and increment answer. ` `        ``i ``=` `pos ``+` `k ` `        ``ans ``+``=` `1` `     `  `    ``return` `ans ` ` `  `# Driver program ` `arr ``=` `[ ``0``, ``1``, ``1``, ``1``, ``1``, ``0` `] ` `n ``=` `len``(arr) ` `k ``=` `2` `print``(minLendings(arr, n, k)) ` ` `  `# This code is contributed by Anant Agarwal. `

## C#

 `// C# program to minimum of number of  ` `// lendings by 1s. ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``static` `int` `minLendings(``int` `[]arr, ``int` `n, ``int` `k) ` `    ``{ ` `         `  `        ``// Create an auxiliary array and fill  ` `        ``// indexes of closest 1s on left side.  ` `        ``int` `[]leftOne = ``new` `int``[n]; ` `        ``int` `oneInd = -1; ` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `            ``if` `(arr[i] == 1) ` `                ``oneInd = i; ` `            ``leftOne[i] = oneInd; ` `        ``} ` `     `  `        ``// Traverse the array from left side. ` `        ``// If current element has a 1 on ` `        ``int` `ans = 0; ` `        ``for` `(``int` `i = 0; i < n;) ` `        ``{ ` `     `  `            ``// Find index of closest 1 after ` `            ``// shift of k-1 or end whichever ` `            ``// is closer. ` `            ``int` `pos = leftOne[Math.Min(i + k - 1, ` `                                          ``n - 1)]; ` `     `  `            ``// If there is no closest 1 within  ` `            ``// allowed range.  ` `            ``if` `(pos == -1 || pos + k <= i) ` `                ``return` `-1; ` `     `  `            ``// If a closest 1 is found, check  ` `            ``// after k jumps and increment answer. ` `            ``i = pos + k; ` `            ``ans++; ` `        ``} ` `         `  `        ``return` `ans; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `[]arr = { 0, 1, 1, 1, 1, 0 }; ` `        ``int` `n = arr.Length; ` `        ``int` `k = 2; ` `         `  `        ``Console.WriteLine(minLendings(arr, n, k)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

Output :

```2
```

The time complexity for this algorithm is O(n) .

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Improved By : vt_m, jit_t

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