Find out the subscript pair of the specified value and solve it in - C language

Find the subscript pair for the specified value

Here are two arrays of integers nums1 and nums2 , Please implement a data structure that supports the following two types of queries ：

 Add up  , Add a positive integer to the  nums2  The subscript corresponding to the element .
 Count  , Statistics satisfy  nums1[i] + nums2[j]  A pair of subscripts equal to the specified value  (i, j)  number （0 <= i < nums1.length  And  0 <= j < nums2.length）.

Realization FindSumPairs class ：

FindSumPairs(int[] nums1, int[] nums2)  Using integer arrays  nums1  and  nums2  initialization  FindSumPairs  object .
void add(int index, int val)  take  val  Add to  nums2[index]  On , namely , perform  nums2[index] += val .
int count(int tot)  Return to satisfaction  nums1[i] + nums2[j] == tot  Subscript pair of  (i, j)  number .

Example ：

Input ：
[“FindSumPairs”, “count”, “add”, “count”, “count”, “add”, “add”, “count”]
[[[1, 1, 2, 2, 2, 3], [1, 4, 5, 2, 5, 4]], [7], [3, 2], [8], [4], [0, 1], [1, 1], [7]]
Output ：
[null, 8, null, 2, 1, null, null, 11]

explain ：
FindSumPairs findSumPairs = new FindSumPairs([1, 1, 2, 2, 2, 3], [1, 4, 5, 2, 5, 4]);
findSumPairs.count(7); // return 8 ; Subscript pair (2,2), (3,2), (4,2), (2,4), (3,4), (4,4) Satisfy 2 + 5 = 7 , Subscript pair (5,1), (5,5) Satisfy 3 + 4 = 7
findSumPairs.add(3, 2); // here nums2 = [1,4,5,4,5,4]
findSumPairs.count(8); // return 2 ; Subscript pair (5,2), (5,4) Satisfy 3 + 5 = 8
findSumPairs.count(4); // return 1 ; Subscript pair (5,0) Satisfy 3 + 1 = 4
findSumPairs.add(0, 1); // here nums2 = [2,4,5,4,5,4]
findSumPairs.add(1, 1); // here nums2 = [2,5,5,4,5,4]
findSumPairs.count(7); // return 11 ; Subscript pair (2,1), (2,2), (2,4), (3,1), (3,2), (3,4), (4,1), (4,2), (4,4) Satisfy 2 + 5 = 7 , Subscript pair (5,3), (5,5) Satisfy 3 + 4 = 7

Our conventional solution to this problem is as follows ：

typedef struct {
    
    int *num1;
    int *num2;
    int num1size;
    int num2size;

} FindSumPairs;


FindSumPairs* findSumPairsCreate(int* nums1, int nums1Size, int* nums2, int nums2Size) {
    

    FindSumPairs *fp=(FindSumPairs *)malloc(sizeof(FindSumPairs));
    int i;
    fp->num1=(int *)malloc(sizeof(int)*nums1Size);
    fp->num2=(int *)malloc(sizeof(int)*nums2Size);
    
    for(i=0;i<nums1Size;i++){
    
        fp->num1[i]=nums1[i];
        
    }
     for(i=0;i<nums2Size;i++){
    
        fp->num2[i]=nums2[i];
        
    }
   
    fp->num1size=nums1Size;
    fp->num2size=nums2Size;
    return fp;
}

void findSumPairsAdd(FindSumPairs* obj, int index, int val) {
    
   obj->num2[index]=obj->num2[index]+val;

}

int findSumPairsCount(FindSumPairs* obj, int tot) {
    
    int i,j;
    int n=0;
    
    for(i=0;i<obj->num1size;i++){
    
       
         for(j=0;j<obj->num2size;j++){
    
             if( obj->num1[i]+obj->num2[j]==tot)
             n++;
         }
    }
    return n;


}

void findSumPairsFree(FindSumPairs* obj) {
    

}

/** * Your FindSumPairs struct will be instantiated and called as such: * FindSumPairs* obj = findSumPairsCreate(nums1, nums1Size, nums2, nums2Size); * findSumPairsAdd(obj, index, val); * int param_2 = findSumPairsCount(obj, tot); * findSumPairsFree(obj); */

Of course, using a hash table is a good way , You can try it , Map our values to the hash table .