Given an array of n elements and an integer K, the task is to find the subarray with minimum value of **||a[i] + a[i + 1] + ……. a[j]| – K|**. In other words, find the contiguous sub-array whose sum of elements shows minimum deviation from K or the subarray whose absolute sum is closest to K.

**Example**

Input:: a[] = {1, 3, 7, 10}, K = 15Output: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 = 6Output:Subarray {1, 2, 3}

The contiguous sub-array [1, 2, 3] shows minimum deviation of 0 from 6.

A **naive approach** would be to check if the sum of each contiguous sub-array and it’s difference from K.

Below is the implementation of the above approach:

`# 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() ` |

*chevron_right*

*filter_none*

**Output:**

-3 5 2 7 6 34

**Time Complexity:** O(N^2)

**Efficient Approach:** If the array only consists of **non-negative integers**, use the sliding window technique to improve the calculation time for sum in each iteration. The sliding window technique reduces the complexity by calculating the new sub-array sum using the previous sub-array sum. Increase the right index till the difference (K-sum) is greater than zero. The first sub-array with negative (K-sum) is considered, and the next sub-array is with left index = i+1(where i is the current right index).

Below is the implementation of the above approach:

`# 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<n): `
` ` ` ` `# Add Last element tp currSum `
` ` `currSum ` `+` `=` `arr[j] `
` ` ` ` `# Save Difference of previous Iteration `
` ` `prevDif ` `=` `currDif `
` ` ` ` `# Calculate new Difference `
` ` `currDif ` `=` `K ` `-` `abs` `(currSum) `
` ` ` ` `# When the Sum exceeds K `
` ` `if` `(currDif <` `=` `0` `): `
` ` `if` `abs` `(currDif) < ` `abs` `(prevDif): `
` ` ` ` `# Current Difference greater in magnitude `
` ` `# Store Temporary Result `
` ` `resultTmp ` `=` `[i, j, currDif] `
` ` `else` `: `
` ` ` ` `# Diffence in Previous was lesser `
` ` `# In previous, Right index = j-1 `
` ` `resultTmp ` `=` `[i, j` `-` `1` `, prevDif] `
` ` ` ` `# In next iteration, Left Index Increases `
` ` `# but Right Index remains the Same `
` ` `# Update currSum and i Accordingly `
` ` `currSum ` `-` `=` `(arr[i]` `+` `arr[j]) `
` ` ` ` `i ` `+` `=` `1`
` ` ` ` `# Case to simply increase Right Index `
` ` `else` `: `
` ` `resultTmp ` `=` `[i, j, currDif] `
` ` `j ` `+` `=` `1`
` ` ` ` `if` `(` `abs` `(resultTmp[` `2` `]) < ` `abs` `(result[` `2` `])): `
` ` `# Check if lesser deviation found `
` ` `result ` `=` `resultTmp `
` ` ` ` `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() ` |

*chevron_right*

*filter_none*

**Output:**

-3 5 2 7 6 34

**Time Complexity:** O(N)

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