Maximum Subarray Sum
Problem Statement :
We define the following: A subarray of array a of length n is a contiguous segment from a[ i ] through a[ j ] where 0 <= i <= j < n. The sum of an array is the sum of its elements. Given an n element array of integers, a, and an integer, m , determine the maximum value of the sum of any of its subarrays modulo m. For example, Assume a = [1, 2, 3 ]and m = 2 . The following table lists all subarrays and their moduli: sum %2 [1] 1 1 [2] 2 0 [3] 3 1 [1,2] 3 1 [2,3] 5 1 [1,2,3] 6 0 The maximum modulus is . Function Description Complete the maximumSum function in the editor below. It should return a long integer that represents the maximum value of subarray sum % m. maximumSum has the following parameter(s): a: an array of long integers, the array to analyze m: a long integer, the modulo divisor Input Format The first line contains an integer q, the number of queries to perform. The next q pairs of lines are as follows: The first line contains two space-separated integers n and (long), m the length of a and the modulo divisor. The second line contains n space-separated long integers a[ i ]. Constraints 2 <= n <= 10^5 1 <= m <= 10^14 1 <= a[ i ] <= 10^18 2 <= the sum of n over all test cases <= 5 * 10^5 Output Format For each query, return the maximum value of subarray sum % m as a long integer. Sample Input 1 5 7 3 3 9 9 5 Sample Output 6
Solution :
Solution in C :
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
typedef unsigned long long ll ;
#define max(x,y) (x ^ ((x ^ y) & -(x < y)))
ll *A,*A2,M;
int intcmp(const void *aa, const void *bb)
{
const ll *a = aa, *b = bb;
return (*a < *b) ? -1 : (*a > *b);
}
ll bisect_left(ll bisect_len,ll x ) {
ll lo = 0;
ll hi = bisect_len;
ll mid = 0;
while ( lo < hi ) {
mid = (lo + hi) / 2;
if ( A2[mid] < x ) lo = mid+1;
else hi = mid;
}
if (lo != bisect_len && A2[lo] == x) {
return x;
}
if (lo == 0) return A2[bisect_len-1];
return A2[lo-1];
}
ll maxCrossingSum(ll l,ll m,ll h) {
ll s0 = 0,maxls = 0,mm=0,r1,r2,bisect_len;
ll i,j;
for ( i = m -1,j=0 ; i >= l && i!= -1; i--,j++ ) {
s0 = (s0 + A[i] ) % M;
if ( maxls < s0 ) maxls = s0;
A2[j] = s0;
}
bisect_len = j;
if ( bisect_len == 0) {
s0 = 0 ;maxls = 0;
for ( i = m ; i <=h ;i++ ) {
s0 = (s0 + A[i] ) % M;
if ( maxls < s0 ) maxls = s0;
}
return maxls;
}
qsort(A2, j, sizeof(ll), intcmp);
s0 = 0;
for ( i = m ; i <= h ; i++) {
s0 = (s0 + A[i] ) % M;
r1 = ( s0 + bisect_left(bisect_len,M-s0-1)) % M;
r2 = ( s0 + maxls ) % M;
mm = max(max(r1,r2),max(s0,mm));
}
return mm;
}
ll maxSum(ll l,ll h) {
ll m,r1,r2,r3;
if ( l == h ) return A[l] % M;
m = ( l + h ) / 2;
r1 = maxCrossingSum(l,m,h);
r2 = maxSum(l,m);
r3 = maxSum(m+1,h);
return max(max(r1,r2),r3);
}
int main() {
ll N,T,i,j,sum;
scanf("%llu",&T);
for (i = 0 ; i < T ; i++ ) {
scanf("%llu",&N);
scanf("%llu",&M);
A = malloc(N * sizeof(ll));
A2 = malloc(N * sizeof(ll));
for ( j = 0 ; j < N ; j++) {
scanf("%llu",&A[j]);
}
sum = maxSum(0,N-1);
printf("%llu\n",sum);
free(A);
free(A2);
}
return 0;
}
Solution in C++ :
In C ++ :
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <set>
using namespace std;
long long mod;
long long arr[100005];
set<long long> f;
int main() {
int T;
scanf("%d", &T);
while (T--) {
int n;
scanf("%d%lld", &n, &mod);
for (int i = 0; i < n; ++i) {
scanf("%lld", &arr[i]);
arr[i] %= mod;
if (i) arr[i] = (arr[i] + arr[i - 1]) % mod;
}
f.clear();
f.insert(0);
set<long long>::iterator it;
long long ans = 0;
for (int i = 0; i < n; ++i) {
it = f.begin();
if (it != f.end())
ans = max(ans, arr[i] - *it);
it = f.upper_bound(arr[i]);
if (it != f.end())
ans = max(ans, (arr[i] - *it + mod) % mod);
f.insert(arr[i]);
}
printf("%lld\n", ans);
}
return 0;
}
Solution in Java :
In Java :
import java.util.*;
public class MaximizeSum {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
while (t-- > 0) {
int n = in.nextInt();
long m = Long.parseLong(in.next());
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
long num = Long.parseLong(in.next()) % m;
if (i == 0)
arr[i] = num;
else
arr[i] = (arr[i -1] + num) % m; // get cumulative sum
}
TreeSet<Long> set = new TreeSet<Long>();
long max = 0;
for (long a : arr) {
if (set.isEmpty()) {
max = a;
set.add(a);
}
else {
max = Math.max(max, a);
Long nextHighest = set.ceiling(a + 1);
if (nextHighest != null)
max = Math.max(max, a - nextHighest + m);
set.add(a);
}
}
System.out.println(max);
}
}
}
Solution in Python :
In Python3 :
import bisect
t = int(input())
for _ in range(t):
n, m = map(int, input().split())
a = list(map(int, input().split()))
csum = [a[0] % m]
for x in a[1:]:
csum.append((csum[-1] + x) % m)
seen = [0]
mx = -1
for s in csum:
idx = bisect.bisect_left(seen, s)
if idx < len(seen):
mx = max(mx, s, (s - seen[idx]) % m)
else:
mx = max(mx, s)
bisect.insort_left(seen, s)
#print(seen)
print(mx)
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