给定一个包含 n 个元素的数组和一个整数 K，任务是找到最小值为||a[i] + a[i + 1] + ……的子数组。 [j]| – K| .换句话说，找到元素总和与 K 的偏差最小的连续子数组或绝对和最接近 K 的子数组。
例子
Input:: a[] = {1, 3, 7, 10}, K = 15
Output: Subarray {7, 10}
The contiguous sub-array [7, 10] shows minimum deviation of 2 from 15.
Input: a[] = {1, 2, 3, 4, 5, 6}, K = 6
Output: Subarray {1, 2, 3}
The contiguous sub-array [1, 2, 3] shows minimum deviation of 0 from 6.
一种天真的方法是检查每个连续子数组的总和及其与 K 的差值。
下面是上述方法的实现：
C++
// C++ code to find sub-array whose
// sum shows the minimum deviation
#include
using namespace std;
int* getSubArray(int arr[], int n, int K)
{
int i = -1;
int j = -1;
int currSum = 0;
// Starting index, ending index,
// Deviation
int* result = new int[3]{ i, j,
abs(K -
abs(currSum)) };
// Iterate i and j to get all subarrays
for(i = 0; i < n; i++)
{
currSum = 0;
for(j = i; j < n; j++)
{
currSum += arr[j];
int currDev = abs(K - abs(currSum));
// Found sub-array with less sum
if (currDev < result[2])
{
result[0] = i;
result[1] = j;
result[2] = currDev;
}
// Exactly same sum
if (currDev == 0)
return result;
}
}
return result;
}
// Driver code
int main()
{
int arr[8] = { 15, -3, 5, 2, 7, 6, 34, -6 };
int n = sizeof(arr) / sizeof(arr[0]);
int K = 50;
// Array to store return values
int* ans = getSubArray(arr, n, K);
if (ans[0] == -1)
{
cout << "The empty array shows "
<< "minimum Deviation";
}
else
{
for(int i = ans[0]; i <= ans[1]; i++)
cout << arr[i] << " ";
}
return 0;
}
// This code is contributed by divyeshrabadiya07 Java
// Java code to find sub-array whose
// sum shows the minimum deviation
class GFG{
public static int[] getSubArray(int[] arr,
int n,int K)
{
int i = -1;
int j = -1;
int currSum = 0;
// Starting index, ending index, Deviation
int [] result = { i, j,
Math.abs(K -
Math.abs(currSum)) };
// Iterate i and j to get all subarrays
for(i = 0; i < n; i++)
{
currSum = 0;
for(j = i; j < n; j++)
{
currSum += arr[j];
int currDev = Math.abs(K -
Math.abs(currSum));
// Found sub-array with less sum
if(currDev < result[2])
{
result[0] = i;
result[1] = j;
result[2] = currDev;
}
// Exactly same sum
if(currDev == 0)
return result;
}
}
return result;
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 15, -3, 5, 2, 7, 6, 34, -6 };
int n = arr.length;
int K = 50;
// Array to store return values
int[] ans = getSubArray(arr, n, K);
if(ans[0] == -1)
{
System.out.println("The empty array " +
"shows minimum Deviation");
}
else
{
for(int i = ans[0]; i <= ans[1]; i++)
System.out.print(arr[i] + " ");
}
}
}
// This code is contributed by dadimadhavPython
# Python Code to find sub-array whose
# sum shows the minimum deviation
def getSubArray(arr, n, K):
i = -1
j = -1
currSum = 0
# starting index, ending index, Deviation
result = [i, j, abs(K-abs(currSum))]
# iterate i and j to get all subarrays
for i in range(0, n):
currSum = 0
for j in range(i, n):
currSum += arr[j]
currDev = abs(K-abs(currSum))
# found sub-array with less sum
if (currDev < result[2]):
result = [i, j, currDev]
# exactly same sum
if (currDev == 0):
return result
return result
# Driver Code
def main():
arr = [15, -3, 5, 2, 7, 6, 34, -6]
n = len(arr)
K = 50
[i, j, minDev] = getSubArray(arr, n, K)
if(i ==-1):
print("The empty array shows minimum Deviation")
return 0
for i in range(i, j + 1):
print arr[i],
main()C#
// C# code to find sub-array whose
// sum shows the minimum deviation
using System;
class GFG{
public static int[] getSubArray(int[] arr,
int n, int K)
{
int i = -1;
int j = -1;
int currSum = 0;
// Starting index, ending index, Deviation
int [] result = { i, j,
Math.Abs(K -
Math.Abs(currSum)) };
// Iterate i and j to get all subarrays
for(i = 0; i < n; i++)
{
currSum = 0;
for(j = i; j < n; j++)
{
currSum += arr[j];
int currDev = Math.Abs(K -
Math.Abs(currSum));
// Found sub-array with less sum
if (currDev < result[2])
{
result[0] = i;
result[1] = j;
result[2] = currDev;
}
// Exactly same sum
if (currDev == 0)
return result;
}
}
return result;
}
// Driver Code
public static void Main(string[] args)
{
int[] arr = { 15, -3, 5, 2, 7, 6, 34, -6 };
int n = arr.Length;
int K = 50;
// Array to store return values
int[] ans = getSubArray(arr, n, K);
if (ans[0] == -1)
{
Console.Write("The empty array " +
"shows minimum Deviation");
}
else
{
for(int i = ans[0]; i <= ans[1]; i++)
Console.Write(arr[i] + " ");
}
}
}
// This code is contributed by rutvik_56Javascript
C++
// C++ code to find non-negative sub-array
// whose sum shows minimum deviation. This
// works only if all elements in array are
// non-negative
#include
using namespace std;
struct Pair
{
int f, s, t;
Pair(int f, int s, int t)
{
this->f = f;
this->s = s;
this->t = t;
}
};
// Function to return the index
Pair* getSubArray(int *arr, int n, int K)
{
int currSum = 0;
int prevDif = 0;
int currDif = 0;
Pair *result = new Pair(
-1, -1, abs(K - abs(currSum)));
Pair *resultTmp = result;
int i = 0;
int j = 0;
while (i<= j && jt) < abs(result->t))
{
// Check if lesser deviation found
result = resultTmp;
}
}
return result;
}
// Driver Code
int main()
{
int arr[] = { 15, -3, 5, 2, 7, 6, 34, -6 };
int n = sizeof(arr) / sizeof(arr[0]);
int K = 50;
Pair *tmp = getSubArray(arr, n, K);
int i = tmp->f;
int j = tmp->s;
int minDev = tmp->t;
if (i == -1)
{
cout << "The empty array shows minimum Deviation"
<< endl;
return 0;
}
for(int k = i + 1; k < j + 1; k++)
{
cout << arr[k] << " ";
}
return 0;
}
// This code is contributed by pratham76 Python
# Python Code to find non-negative
# sub-array whose sum shows minimum deviation
# This works only if all elements
# in array are non-negative
# function to return the index
def getSubArray(arr, n, K):
currSum = 0
prevDif = 0
currDif = 0
result = [-1, -1, abs(K-abs(currSum))]
resultTmp = result
i = 0
j = 0
while(i<= j and j输出：
-3 5 2 7 6 34时间复杂度： O(N^2)
高效的方法：如果数组只由非负整数组成，则使用滑动窗口技术来提高每次迭代求和的计算时间。滑动窗口技术通过使用先前的子阵列总和计算新的子阵列总和来降低复杂度。增加正确的索引，直到差异（K-sum）大于零。考虑第一个具有负数（K-sum）的子数组，下一个子数组的左索引 = i+1（其中 i 是当前的右索引）。
下面是上述方法的实现：
C++
// C++ code to find non-negative sub-array
// whose sum shows minimum deviation. This
// works only if all elements in array are
// non-negative
#include
using namespace std;
struct Pair
{
int f, s, t;
Pair(int f, int s, int t)
{
this->f = f;
this->s = s;
this->t = t;
}
};
// Function to return the index
Pair* getSubArray(int *arr, int n, int K)
{
int currSum = 0;
int prevDif = 0;
int currDif = 0;
Pair *result = new Pair(
-1, -1, abs(K - abs(currSum)));
Pair *resultTmp = result;
int i = 0;
int j = 0;
while (i<= j && jt) < abs(result->t))
{
// Check if lesser deviation found
result = resultTmp;
}
}
return result;
}
// Driver Code
int main()
{
int arr[] = { 15, -3, 5, 2, 7, 6, 34, -6 };
int n = sizeof(arr) / sizeof(arr[0]);
int K = 50;
Pair *tmp = getSubArray(arr, n, K);
int i = tmp->f;
int j = tmp->s;
int minDev = tmp->t;
if (i == -1)
{
cout << "The empty array shows minimum Deviation"
<< endl;
return 0;
}
for(int k = i + 1; k < j + 1; k++)
{
cout << arr[k] << " ";
}
return 0;
}
// This code is contributed by pratham76
Python
# Python Code to find non-negative
# sub-array whose sum shows minimum deviation
# This works only if all elements
# in array are non-negative
# function to return the index
def getSubArray(arr, n, K):
currSum = 0
prevDif = 0
currDif = 0
result = [-1, -1, abs(K-abs(currSum))]
resultTmp = result
i = 0
j = 0
while(i<= j and j输出
-3 5 2 7 6 34时间复杂度： O(N)
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