दो क्रमबद्ध सरणियों से असामान्य तत्व प्रिंट करें

दो क्रमबद्ध सरणियां दी गई हैं और आउटपुट में उनके असामान्य तत्व प्रदर्शित होने चाहिए

Given : array1[]= {1, 4, 6, 9, 12}     array2[]= {2, 4, 7, 8, 9, 10} Output : 1 2 6 7 8 10 12

एल्गोरिदम

START Step 1 -> declare two arrays array1 and array2 with elements as int and variables n1, n2, i to 0 and j to 0 Step 2 -> calculate number of elements in array1 sizeof(array1)/sizeof(array1[0]) Step 3-> calculate number of elements in array2 sizeof(array2)/sizeof(array2[0]) Step 4 -> Loop While till i array2[j]         Print array2[j++ ]         End ELSE IF     ELSE         i++ and j++     End ELSE Step 5 -> End Loop While Step 6 -> loop While i < n1 && array1[i]!=array2[j]     Print array1[i++] Step 7 -> End Loop While Step 8 -> loop While j < n2 && array2[j]!=array1[i]     Print array2[j++] Step 9 -> End Loop While STOP

उदाहरण

#include  int main(int argc, char const *argv[]) {     int array1[]= {1, 4, 6, 9, 12};     int array2[]= {2, 4, 7, 8, 9, 10};     int n1, n2, i=0, j=0;     n1 = sizeof(array1)/sizeof(array1[0]); //Calculating number of elements in array1     n2 = sizeof(array2)/sizeof(array2[0]); //Calculating number of elements in array2     while(i < n1 && j < n2) {         if(array1[i]  array2[j]) //checking whether the element of array2 is smaller than array1             printf("%d\n", array2[j++]);         else { //if they are equal increment both i and j             i++;             j++;         }     }     while(i < n1 && array1[i]!=array2[j]) //print remaining array1         printf("%d\n", array1[i++]);     while(j < n2 && array2[j]!=array1[i]) //print remaining array1         printf("%d\n", array2[j++]);     return 0; }

आउटपुट

यदि हम उपरोक्त प्रोग्राम चलाते हैं तो यह निम्नलिखित आउटपुट उत्पन्न करेगा

1 2 6 7 8 10 12