can someone tell me the approach of the question like when we do recursive call where do we check the condition
#dsa help
silent wing
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runic marten
can you send the full image
silent wing
silent wing
can you send the full image
here you go 🙂
molten knoll
can someone tell me the approach of the question like when we do recursive call ...
Approach it like this: for every place in the array check the following places for the condition
You will need 2 for loops, nested into each other
silent wing
i got da answer guys
idle tokenBOT
"""class Solution {
static int count = 0;
public int reversePairs(int[] nums) {
count = 0;
mergeSort(nums);
return count;
}
public void mergeSort(int[] arr){
int n = arr.length;
if(n <= 1){
return;
}
int[] a = new int[n/2];
int[] b = new int[n-n/2];
int k = 0;
for(int i = 0; i < a.length; i++) a[i] = arr[k++];
for(int j = 0; j < b.length; j++) b[j] = arr[k++];
mergeSort(a);
mergeSort(b);
merge(arr,a,b);
}
public void merge(int[] target, int[] a, int[] b){
int i = 0, j = 0;
while(i < a.length){
while(j < b.length && a[i] > 2L * b[j]){
j++;
}
count += j;
i++;
}
i = 0;
j = 0;
int k = 0;
while(i < a.length && j < b.length){
if(a[i] < b[j]){
target[k++] = a[i++];
}
else{
target[k++] = b [j++];
}
}
while(i < a.length) target[k++] = a[i++];
while(j < b.length) target[k++] = b[j++];
}
}""" ```