Subsequence closest to 0

Given the integers m, n and array X [N] to find an I, make X [I] + X [I + 1] + X [I + 2] + X [I + 3] + X [I + 4]… X [I + M] is closest to zero.

(0 <= I <n-m)

I. Brute Force Solution

Traverse each I value, calculate the sum of sub-sequences, and then find

Int find (int A [], int N, int m) // search for m + 1 numbers to minimize their sum {int I = 0; int thissum = 0; int J = 0; int ans = int_max; for (I = 0; I <n-m; I ++) // time complexity is O (M * n) {thissum = 0; For (j = I; j <= I + m; j ++) thissum + = A [J]; If (ABS (INT) (thissum) <ABS (INT) (ANS) ans = ABS (INT) thissum);} return ans ;}

Obviously, the time complexity is O (M * n), and the space complexity is O (1)

Ii. Dynamic Planning:

Create an array DP []. DP [I] Stores X [I] + X [I + 1] +... For the value of X [I + M], the recursive relationship is:

DP [I + 1] = DP [I]-A [I] + A [I + 1 + M], traverse the DP [] array, and find the nearest zero.

int find2(int a[],int n,int m){int *dp=(int *)malloc(n*sizeof(int));int i=0;int tmp=0;for(i=0;i<=m;i++){tmp+=a[i];}dp[0]=tmp;for(i=1;i<n-m;i++){dp[i]=dp[i-1]+a[i+m]-a[i-1];}int ans=INT_MAX;for(i=0;i<n;i++){if(abs((int)dp[i])<ans)ans=abs((int)dp[i]);}return ans;}

The time complexity is O (n), and the space complexity is O (n)

You can also make some improvements to the above Code, the above DP [] array, just use a variable.

int find3(int a[],int n,int m)   {int tmp=0;int ans=INT_MAX;int i=0;for(i=0;i<=m;i++)tmp+=a[i];if(abs((int)tmp)<ans)ans=abs((int)tmp);for(i=1;i<n-m;i++){tmp=tmp-a[i-1]+a[i+m];if(abs((int)tmp)<ans)ans=abs((int)tmp);}return ans;}

The time complexity is O (n), and the space complexity is O (1)

Complete code:

# Include <iostream> # include <cstdlib> # include <cmath> # include <ctype. h> using namespace STD; int find (int A [], int N, int m) // find m + 1 numbers to minimize their sum {int I = 0; int thissum = 0; Int J = 0; int ans = int_max; for (I = 0; I <n-m; I ++) // The time complexity is O (M * n) {thissum = 0; For (j = I; j <= I + m; j ++) thissum + = A [J]; If (ABS (INT) (thissum) <ABS (INT) (ANS) ans = ABS (INT) thissum);} return ans;} int find2 (int A [], int N, int m) {int * dp = (int *) malloc (N * sizeof (INT); int I = 0; int TMP = 0; for (I = 0; I <= m; I ++) {TMP + = A [I];} DP [0] = TMP; for (I = 1; I <n-m; I ++) {DP [I] = DP [I-1] + A [I + M]-A [I-1];} int ans = int_max; for (I = 0; I <N; I ++) {If (ABS (INT) DP [I]) <ans) ans = ABS (INT) DP [I]);} return ans ;} int find3 (int A [], int N, int m) // {int TMP = 0; int ans = int_max; int I = 0; for (I = 0; I <= m; I ++) TMP + = A [I]; If (ABS (INT) TMP) <ans) ans = ABS (INT) TMP ); for (I = 1; I <n-m; I ++) {TMP = TMP-A [I-1] + A [I + M]; If (ABS (INT) TMP) <ans) ans = ABS (INT) TMP);} return ans;} int main () {int A [10] = {1,-2, 3,-4, -5,-6, 7, 0}; cout <find (A,) <Endl; cout <find2 (A,) <Endl; cout <find3 (A, 10, 3) <Endl; System ("pause"); Return 0 ;}