count += remainderFrequency[0] * (remainderFrequency[0] - 1) / 2;

• Java arithmetic operators are executed left to right
• Fail test case in leetcode: an array length of 60000 containing only 60.
• Array: [60 60 60 60 60 60 60 ....] (60000 60s are in this input array)
• remainderFrequency[0] * remainderFrequency[0] - 1 = 60000 * 59999 = 3599940000
• 3599940000 is larger than the largest integer value in Java causing overflow
• Solution: casting one of the number to long to preserve precision and prevent integer overflow

count += (long)remainderFrequency[0] * (remainderFrequency[0] - 1) / 2;