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public int solve(int[] A) {
int n = A.length;
int mod = 1000000007;
// calculate prefix sum and suffix sum arrays
long[] prefixSum = new long[n];
long[] suffixSum = new long[n];
prefixSum[0] = A[0];
suffixSum[n-1] = A[n-1];
for (int i = 1; i < n; i++) {
prefixSum[i] = prefixSum[i-1] + A[i];
suffixSum[n-1-i] = suffixSum[n-i] + A[n-1-i];
}
// calculate the maximum sum of one subarray from the left
long[] maxSumLeft = new long[n];
long curSum = 0;
long curMax = Long.MIN_VALUE;
for (int i = 0; i < n; i++) {
curSum += A[i];
curMax = Math.max(curMax, curSum);
maxSumLeft[i] = curMax;
curSum = Math.max(curSum, 0);
}
// calculate the maximum sum of one subarray from the right
long[] maxSumRight = new long[n];
curSum = 0;
curMax = Long.MIN_VALUE;
for (int i = n-1; i >= 0; i--) {
curSum += A[i];
curMax = Math.max(curMax, curSum);
maxSumRight[i] = curMax;
curSum = Math.max(curSum, 0);
}
// calculate the maximum sum of three subarrays
long maxSum = Long.MIN_VALUE;
for (int i = 1; i < n-1; i++) {
long sum1 = maxSumLeft[i-1];
long sum2 = maxSumRight[i+1];
long sum3 = prefixSum[i-1] + suffixSum[i+1];
maxSum = Math.max(maxSum, sum1+sum2+sum3);
}
return (int)(maxSum % mod);
}
In python def max_sum_subarrays(arr):
n = len(arr)
k = n // 3 # size of each window
mod = 10**3 + 7 # modulo value
# initialize windows
window1 = arr[:k]
window2 = arr[k:2*k]
window3 = arr[2*k:3*k]
# initialize sums
sum1 = sum(window1)
sum2 = sum(window2)
sum3 = sum(window3)
# iterate over the rest of the array and update windows and sums
for i in range(k, n-k):
sum1 += arr[i] - window1[i-k]
sum2 += arr[i+k] - window2[i-k]
sum3 += arr[i+2*k] - window3[i-k]
# update windows
window1[i-k] = arr[i]
window2[i-k] = arr[i+k]
window3[i-k] = arr[i+2*k]
# update maximum sums
sum1 = max(sum1, 0)
sum2 = max(sum2, 0)
sum3 = max(sum3, 0)
# return the positive remainder after dividing the sum with modulo value
return (sum1 + sum2 + sum3) % mod
public int solve(int[] A) {
int n = A.length;
int mod = 1000000007;
// calculate prefix sum and suffix sum arrays
long[] prefixSum = new long[n];
long[] suffixSum = new long[n];
prefixSum[0] = A[0];
suffixSum[n-1] = A[n-1];
for (int i = 1; i < n; i++) {
prefixSum[i] = prefixSum[i-1] + A[i];
suffixSum[n-1-i] = suffixSum[n-i] + A[n-1-i];
}
// calculate the maximum sum of one subarray from the left
long[] maxSumLeft = new long[n];
long curSum = 0;
long curMax = Long.MIN_VALUE;
for (int i = 0; i < n; i++) {
curSum += A[i];
curMax = Math.max(curMax, curSum);
maxSumLeft[i] = curMax;
curSum = Math.max(curSum, 0);
}
// calculate the maximum sum of one subarray from the right
long[] maxSumRight = new long[n];
curSum = 0;
curMax = Long.MIN_VALUE;
for (int i = n-1; i >= 0; i--) {
curSum += A[i];
curMax = Math.max(curMax, curSum);
maxSumRight[i] = curMax;
curSum = Math.max(curSum, 0);
}
// calculate the maximum sum of three subarrays
long maxSum = Long.MIN_VALUE;
for (int i = 1; i < n-1; i++) {
long sum1 = maxSumLeft[i-1];
long sum2 = maxSumRight[i+1];
long sum3 = prefixSum[i-1] + suffixSum[i+1];
maxSum = Math.max(maxSum, sum1+sum2+sum3);
}
return (int)(maxSum % mod);
}
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